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* BY THE SAME AUTHOR
ELEMENTARY ALGEBRA.
FOURTH EDITION. REVISED AND ENLARGED.
Globe 8vo. 4«. 6d. Key, 10& 6c?.
This book has been thoroughly revised and the early chapters
remodelled and simplified; the nmnber of examples has been veiy
greatly increased; and chapters on Logaritiims and Scales of Notation
have been added.
AN ELEMENTARY TREATISE
ON CONIC SECTIONS.
TENTH EDITION.
Crown 8vo. 7«. M. Key, 10«. 6d.
AN ELEMENTARY TREATISE
ON SOLID GEOMETRY.
THIRD EDITION.
Crown 8vo. 9$. 6d.
SOLUTIONS OF THE EXAMPLES
IN
A TREATISE ON ALGEBRA.
SECOND EDITION.
Crown 8vo. 10«. 6d.
Sonbon: MACMILLAN AND CO.
A TREATISE
ON
ALGEBEA
A TEBATISE
ON
ALGEBEA
BY
CHARLES SMITH, M.A.
MABTBB 07 SIDNEY BUSSBX OOLLSGE, OAMBBIDGB.
THIRD EDITION.
Uonlfon:
MACMILLAN AND CO.
AND NEW YOBK.
1892
[The Right of Translation is reserved.]
4^
b w »
•> fc W W V ^
* * V V fc^ »
First Edition, 1888.
Second Edition, 1890.
2'Atrd Edition, with additions, 1892.
€DUCATION DEPT
PEEFACE TO THE FIRST EDITION.
The following work is designed for the use of the
higher classes of Schools and the junior students in the
Universities. Although the book is complete in itself, in
the sense that it begins at the beginning, it is expected
that students who use it will have previously read some
more elementary work on Algebra : the simpler parts of
the subject are therefore treated somewhat briefly.
I have ventured to make one important change from
the usual order adopted in English textbooks on Algebra,
namely by considering some of the tests of the conver
gency of infinite series before making any use of such
series : this change will, I feel sure, be generally approved.
The order in which the different chapters of the book may
be read is, however, to a great extent optional.
A knowledge of the elementary properties of Deter
minants is of great and increasing practical utility ; and
I have therefore introduced a short discussion of their
fundamental properties, founded on the Treatises of Dostor
and Muir.
No pains have been spared to ensure variety and inte
rest in the examples. With this end in view, hundreds
of examination papers have been congultd; including, with
VI MEFACK
very few exceptions, every paper which has been set in
Cambridge for many years past. Amongst the examples
will also be found many interesting theorems which have
been taken from the different Mathematical Journals.
I am indebted to many friends for their kindness in
looking over the proofsheets, for help in the verification
of the examples, and for valuable suggestions. My especial
thanks are due to the following members of Sidney Sussex
College: Mr S. R. Wilson, M.A., Mr J. Edwards, M.A.,
Mr S. L. Loney, M.A., and Mr J. Owen, B.A.
CHAELES SMITH.
Cambbibob,
December 12th, 1887.
PREFACE TO THE THIRD EDITION.
A Chapter on Theory of Equations has been added,
which it is hoped will increase the value of the book.
CONTENTS.
CHAPTER I.
PAOB
Defimtions 1
CHAPTER II.
Fundamental Laws.
Negative qttantities 9
Addition of terms 10
Subtraction of terms 11
Multiplication of monomial expressions 14
Law of Signs 15
The factors of a product may be taken in any order ... 16
Fondamental Index Law 18
rHyifdon of monomial expressions 19
Multinomial expressions 22
Commutatiye Law, Distributive Law and Associative Law . . 23
CHAPTER III
Addition. Subtraction. Brackets.
Addition of any multinomial expressions 26
Subtraction of multinomial expressions 27
Brackets 28
Examples L 29
• • •
VIU CONTENTS.
CHAPTER IV.
Multiplication. page
Product of two mnltinomial expressions 32
Detached coefficients 86
Square of a multinomial expression 39
Continued products 40
Examples II 42
CHAPTER V.
Division.
Division by a multinomial expression 46
Extended definition of division 60
Examples HI. 51
CHAPTER VI.
Factors.
Monomial factors . . > 53
Factors found by comparing with known identities ... 54
Factors of quadratic expressions found by inspection ... 55
Examples IV. 56
Factors of general quadratic expression . " 67
Factors found by rearrangement and grouping of terms . . . 60
Examples V • .64
Divisibility of x**± a** by a; ±rt . . .... ... .65
Remainder Theorem 66
An expression of the nth degree cannot vanish for more than n
values of x, unless it vanishes for all values of a; . . .69
Cyclical order 71
Symmetrical expressions 71
Factors found by use of Remainder Theorem 72
Examples VI 73
CHAPTER VII.
Highest Common Factor. Lowest Common Multiple.
Monomial common factors 76
Multinomial common factors 77
Examples Vn. 83
Lowest common multiple 83
Examples VIII .86
XJONTENXa IX
CHAPTER VIII.
Fracjtions.
PAGE
A fraction is not altered by multiplying its nmnerator ^nd denomi
nator by the same quantity 87
Reduction of fractions to a common denominator .... 89
Addition and subtraction of fractions 90
Multiplication and division of fractions 92
Important theorems concerning fractions formed from given fractions 95
Examples IX. ..." 98
* ^ *
CHAPTER IX.
Equations. One Unknown Quantity.
General principles applicable to all equations 105
Simple equations . . . . 106
Special forms of simple equations 107
The problem of solving an equation the same as the problem of
finding the lactors of an expression . . . : '. . 109
Quadratic equations 109
Discussion of roots of a quadratic equation 112
Zero and infinite roots 113
Equations not integral 115
Irrational equations 118
A quadratic equation can only have two roots . . . . . 120
Relations between the roots and the coefficients of. a quadratic
equation . 122
Relations between the roots and the coefficients of any equation . 123
Equations with given roots . 124
Discussion of possible values of a trinomial expression . . . 126
Examples X 129
Equations of higher degree than the second 134
Equations of the same form as quadratic equations . . . 134
Reciprocal equations .>.•.' 136
Roots found by inspection 137
Binomial equations 138
Chibe roots of unity 139
Examples XI .140
X CONTENTS.
CHAPTER X.
Simultaneous Equations.
Equations of the first degree with two unknown quantities . . 144
Discussion of solution 148
Equations of the first degree with three unknown quantities . . 149
Method of undetermined multipliers 150
Equations with more than three unknown quantities . . . 153
Examples Xn 154
Simultaneous equations of the second degree 156
Examples XTTT 162
Equations with more than two unknown quantities .  . . 164
Examples XIV. 169
CHAPTER XL
Pboblemb.
Problems not always satisfied by the solutions of the corresponding
equations ' 172
Examples XV ^ , • 176
CHAPTER XIL
Miscellaneous Theorems and Examples.
Examples of elimination 181
Equations with restrictions on the values of the letters . a . 185
Identities deduced from tho factors of a'* +6* + c83«i/c . . . 186
Examples XVI 189
CHAPTER XIII.
Powers and Roots. Fractional and Nbgattvb Indiobs.
\ Index Laws 199
I Boots of arithmetical numbers 201
Surds obey the Fundamental Laws of Algebra .... 202
Fractional and negative indices 204
Rationalizing factors 209
Examples XVII. 211
i
CONTENTS. XI
CHAPTER XIV.
Surds. Imaginary and Complex Quantities.
PAOB
Properties of Surds 213216
If a+^b=c+^df where a and c toe rational and ^b and Jd are
irrational, then a =sc and 6= <2 216
If either of two conjugate quadratic surdfl is a factor of a rational
expression, so also in the other 216
Squareroot of a+V^ 217
Examples XVm 219
Imaginary and complex quantities 220
Complex quantities obey the Fundamental Laws of Algebra . . 222
Definition and properties of the modulus of a complex quantity 223225
If either of two conjugate complex qnantitie<i in a factor of a real
expression, so also is the other 225
CHAPTER XV.
Square and Cube Roots.
Square roots found by inspection 228
Square root of any algebraical expression 229
Square root found by equating coefficients 230
Extended definition of Square root 231
When any number of terms of a square root have been found, as
many more terms can be found by ordinary division . . 232
When n figures of a square root have been found by the ordinary
method, n2 more figures can always be found by division . 233
Cube root 234
Method of finding the nth root of any algebraical expression . . 237
Examples XIX 237
CHAPTER XVI.
Ratio and Proportion.
Ratio. Compound ratio. Duplicate ratio 240
A ratio is made more nearly equal to unity by adding;: the same
positive quantity to each of its terms 241
Incommensurable numbers 242
Proportion 243
Continued proportion. Mean proportional 244
Geometrical and algebraical definitions compared .... 245
Xll CONTENTS.
PAQE
Variation 247
Indeterminate Forms . 250
Examples XX 252
CHAPTER XVII.
The Progressions.
Arithmetical progression 254
Geometrical progression 259
Harmonical progression . . . 264
Examples XXI. . . . 266
CHAPTER XVIII.
Systems of Numeration.
Expression of any integer in any scale of notation .... 270
Radix fractions . . 272
The difference between any number (expressed in the scale r) and
the'sum of its digits is diyisible by r  1 274
Rule for casting out the nines 275
Examples XXn 276
CHAPTER XIX.
Permutations and Combinations.
Permutations of different things 279
Permutations all together of things which are not all different . 280
Combinations 282
vPr—ffinT '. • • ^^
Greatest value of „6V 286
(x+y)^i»=x^i»+x^»i*»^i+'+v^n 286
Vandermonde's Theorem [see also p. 309] 287
Homogeneous products [see also p. 351] 288
Examples XX TH 293
CHAPTER XX.
The Binomial Theorem.
Proof of the binomial theorem for a positive integral exponent . 296
Proof by induction 297
Greatest term 300
Examples XXIV 302
Properties of the poefficients of a binomial expansion . . . 304
• • •
CONTENTS, XUl
PAGE
Continued product of n binomial factors of the form x+a, x + b, &c. 307
Vandermonde's Theorem [see also p. 287] 309
The multinomial theorem 310
Examples XXV 313
CHAPTER XXL
CONVERGENCT AND DIVERGENCY OF SERIES.
Convergency and divergency of series, all of whose terms havfe the
same sign 317325
Series whose terms are alternately positive and negative . . . 325
Application to the Binomial, the Exponential and the Logarithmic
Series 326328
Product of convergent series 328
If 2a^*'=26^'', for all values of x for which the series are con
vergent, then ay= 6^ 330
Examples XXVL . 331
CHAPTER XXII.
The Binomial Theorem. Any Index.
Proof of the theorem 334
Euler's proof 335
Greatest term 339
Examples XXVII 340
Sum of the first r + 1 coefficients of Aq + OiX + a^x^ + 342
Binomial Series 343
Expansion of multinomials 347
Combinations and Permutations with repetitions .... 349
Homogeneous products [see also p. 288] 351
Examples XXVIII 354
CHAPTER XXIII.
Partial Fractions. Indeterminate Coefficients.
Decomposition into partial fractions ...... 361
Case of imaginary factors . 364
Case of equal factors • • 365
Indeterminate coefficients 367
Examples XXIX 368
•
XIV CONTENTS.
CHAPTER XXIV.
Exponential Theorem. Logarithms. Logarithmic Series.
PAGB
The Ezi>onential Theorem 372
Examples XXX 877
Properties of logarithms 879
The logarithmic series 880
Ganchy's theorem 882
Series for calculating logarithms 882
Examples XXXL 884
Common logarithms 887
Compound interest and annuities 390
Examples XXXIL 392
CHAPTER XXV.
Summation op Series.
Sum of series found by expressing t*^ in the form v^  v^_ . . 394
Sum of series whose general term is
{(a+n1.6)(a+n6)...(a+n+r2.6)} . 396
Sum of series whose general term is
l/{(a+ n^ . b) (a+nb)...{a+ n+r2. b)} . 399
Sum of squares and sum of cubes of the first n numbers . . 402
SumofF+ 2^ + ...+?:»• 408
Piles of shot 404
Figurate Numbers 405
Polygonal Numbers 406
Examples XXXin 406
Sum of series whose general term is — ^ ^ ' — ^^ — ' 409
b(b + x){b + 2x)...{b + nl . x)
XojfiuP, where ap=A^+AriP^^ + .,. + AQ 411
Series whose law is not given 418
Method of Differences 414
Recurring Series 416
Convergenoy of Infinite products 422
n jt'i.* r «i . . m(iiil) , m(iiil)(m2) .«»
Conditions for convergenoy of 1 tfe m + A: — jr^ ± ^ ^ ^ ^ + . . . 428
Zu„ and "^^^ both convergent or both divergent .... 425
CONTENXa XV
PAGE
2u^ oonvergent or divergent according as the limit of
l!!*t«^>l . 427
Examples XXXIY. 428
•(
CHAPTER XXVL
Inequalities.
Elementary Principles 434
Product of any given number of positiye quantities, whose sum is
given, is greatest when the quantities are equal . . . 435
The arithmetic mean of any number of positive quantities is greater
than their geometric mean 436
The sum of any given number of positive quantities, whose product
is given, is least when the quantities are all equal . • . 437
If m, a, /?,... ore positive and m = a + /3 + ..., then
n n n
2a» 2a) *
> j — y , unless « is a positive fraction .... 441
Examples XXXV 412
CHAPTER XXVII.
Continued Fractions.
Convergents ofa+... are alternately less and greater than
the continued fraction 446
Law of formation of successive conveigeuts 446
Reduction of any rational fraction to a continued fraction . 448
Properties of convergents • 449
Examples XXXVL 453
General convergent 455
Periodic continued fractions 457
Gonveigency of continued fractions 459
Reduction of quadratic surds to continued fractions . . . 454
Series expressed as continued fractions . .^ . . . . 469
Examples XXXVII 472
XVI CONTENTS.
CHAPTER XXVIII.
Theory of Numbers
tAGB
The Sieve of Eratosthenes 478
Properties of primes 479481
Highest power of a prime contained in n 482
The product of any n consecutive integers is divisible by (u . . 482
Fermat*8 Theorem [see also p. 493] 484
Number of divisors of a given number 485
Number of positive integers less than a given number and prime to
it [see also p. 495] 486
Forms of square numbers 488
Examples XXXVIIL 489
Congruences 491
Wilson's Theorem 493
Extension of Permat's Theorem 496
Lagrange's Theorem 497
Reduction of Fractions to Circulating Decimals .... 498
Examples XXXIX. 500
CHAPTER XXIX.
Indeterminate Equations.
Integral solutions of ax^by=c can always be found if a and b are
prime to one another 503
General solution of ax by = Cf having given one solution . . 503
General solution of a^ + 6i/=c, having given one solution . . 504
Number of positive integral solutions of ox + &^=c . . . . 505
Integral solutions ot ax + by + cz = d, a'x + b'y + c'z = d' . . . 507
Examples XL 509
* • t c
CHAPTER XXX
Probability.
Definition of probability . 511
Exclusive Events . . . 513
Independent Events  . . . 514
Dependent Events  . 515
qONTENTS. XVll
PAQB
Jhrobal^ility of ab eVent happening r times in n trials . . 517
Ezpeotstion 519
Inverse Probability 620
Probability of Testimony 522
Examples XLL 525
. GHAPTEE XX^XI..
Determinants.
Definition uid properties of determinants .... 529541
Mnltiplication of determinants 542
Simultaneous Equations of the First Degree 544
Elimination 546
Sylvester's method of Elimination 547
Examples XLII 548
CHAPTER XXXn.
Theory op Equations.
Every equation of the nth degree has n roots . . • . • 552
Belations between the roots and the coefficients of an equation . 553
Sum of the mth powers of the roots 555
Symmetrical functions of roots 556
Transformation of Equations 557
Beciprocal Equations 559
Examples XTiTTT 562
Imaginary and quadratic surd roots occur in pairs . . . 563
Boots common to two equations 565
Hoots connected by any given relation 565
Commensurable Boots 566
Examples XLIV 567
Derived Functions 569
Equal Boots 570
Continuity of a rational integral function 572
If /(a) and/()S) are of contrary signs, a real root of /(a:)=0 lies
between a and /3 ^73
The Discriminating Cubic ^74
Bolle's Theorem ^75
XVlll CONTENTS.
PAQlt
Descartes' Boles of Signs •*•.««•• 576
Examples XL V. \ 678
Cubic Equations 580
Biquadratic Equations 581
Storm's Theorem 583
Synthetic Diyision 587
Homer's Method of approximating tu the real roots of any equation 590
Examples XLVI 593
Answers to the Examples 596
«
CHAPTER I.
Definitions.
1. Algebra, like Arithmetic, is a science which treats
of numbers.
In Arithmetic numbers are represented by figures
which have determinate values. In Algebra the letters of
the alphabet are used to represent numbers, and each
letter can stand for any number whatever, except that in
any connected series of operations each letter must through
out be supposed to represent the same number.
Since the letters employed in Algebra represent any
numbers whatever, the results arrived at must be equally
true of all numbers.
2. The numbers treated of may be either whole
numbers or fractions.
All concrete quantities such as values, lengths, areas,
periods of time, &c., with which we have to do in Algebra,
must be measured by the number of times each contains
some unit of its own kind. Thus we have lengths of 4, f ,
5\, the unit being an inch, a yard, a mile, or any other
fixed length. It is only these numbers with which we are
concerned, and our symbols of quantity, whether figures or
letters, always represent numbers. On this account the
word qua/ntity is often used instead of number »
3. The sign + , which is read ' plus,' is placed before a
number to indicate that it is to be added to what has gone
s. A. 1
2 DEFINITIONS.
before!: '^^Sius ^ftf 8 Z^ejiife that 3 is to be added to 6;
6+.3;f;2.i](j9ans^*^tKal S 7s fo be added to 6 and then 2
acJjdfejJ \o^p^^ f^uft; ^q^also'ji + b means that the number
whicli is represented'by'ft fe* to be added to the number
which is represented by a ; or, expressed more briefly, it
means that 6 is to be added to a ; again a + b + c means
that b is to be added to a and then c added to the
result.
4. The sign — , which is read 'minus/ is placed before
a number to indicate that it is to be subtracted from what
has gone before. Thus a— 6 means that 6 is to be subtracted
from a; a — b — c means that b is to be subtracted from a,
and then c subtracted from the result; and a — 6 + c means
that 6 is to be subtracted from a, and then c added to the
result.
Thus in additions and subtractions the order of the
operations is from left to right.
5. The sign x , which is read 'into/ is placed between
two numbers to indicate that the first number is to be
multiplied by the second. Thus axb means that a is to
be multiplied by b; also axbxc means that a is to be
multiplied by b, and the result multiplied by c.
The sign x is however generally omitted between two
letters, or between a figure and a letter, and the letters
are placed consecutively. Thus ah means the same as
axb, and 5ab the same as 5 x a x 6.
The sign of multiplication cannot be omitted between
figures : 63 for example does not stand for 6x3 but for
sixtythree, as in Arithmetic.
Sometimes the x is replaced by a point, which is
placed on the line, to distinguish it from the decimal
point which is placed above the line. Thus axbxc,
a.b.c and abc all mean the same, namely that a is to
be multiplied by b and the result multiplied by c.
6. The sign ^, which is read 'divided by' or *by/
is placed between two numbers to indicate that the first
DEFINITIONS. 3
number, called the dividend, is to be divided by the
second number, called the divisor. Thus aib means
that a is to be divided by b ; also arbic means that
a is to be divided by 6, and the result divided by c;
and aib xc means that a is to be divided by b and the
result multiplied by c.
Thus in multiplications and divisions the order of the
operations is from left to right.
7. When two or more numbers are multiplied together
the result is called the continued product, or simply the
product; and each number is called a fdctor of the
product.
When the factors are considered as divided into two
sets, each is called the coefficient, that is the cofactor
of the other. Thus in Sa&rr, 3 is the coefficient of abx,
3a is the coefficient of bx, and 3a6 is the coefficient of x.
When one of the factors of a product is a number
expressed in figures, it is called the numerical coefficient
of the product of the other factors.
8. When a product consists of the same factor
repeated any number of times it is called a power of that
factor. Thus aa is called the second power of a, aaa is
called the third power of a, aooa is called the fourth
power of a, and so on. Sometimes a is called the first
power of a.
Special names are also given to a^ and to aaa ; they
are called respectively the square and the cube of a.
9. Instead of writing aa, aaa, &c., a more convenient
notation is adopted as follows : a' is used instead of aa,
a' is used instead of aaa, and a" is used instead of
aaaa , the factor a being taken n times; the small
figure placed above and to the right of a shewing the
number of times the factor a is to be taken. So also
aW is written instead of aaM), and similarly in other
cases.
The small figure, or letter, placed above a symbol to
1—2
4 1>EFIN1TI0NS.
indicate the number of times that symbol is to be taken
as a factor is called the index or the exponent Thus a**
means that the factor a is to be taken n times, or that
the nth power of a is to be taken, and n is called the
index.
When the factor a is only to be taken once, we do
not write it a\ but simply a,
10. A number which when squared is equal to any
number a is called a square root of a, and is represented
by the symbol ^a, or more often by *Ja : thus 2 is V^*
since 2' = 4.
A number which when cubed is equal to any number
a is called a cvhe root of a, and is represented by the
symbol ^a : thus 3 is ^27, since 3' = 27.
In general, a number which when raised to the nth.
power, where n is any whole number, is equal to a, is
called an nth root of a, and is represented by the
symbol ^a.
The sign V was originally the initial letter of the
word radix. It is often called the radical sign.
11. A root which cannot be obtained exactly is
called a surd, or an irrational quantity: thus V^ ^^^
i/4s are surds.
The approximate value of a surd, for example of \/7,
can be found, to any degree of accuracy which may be
desired, by the ordinary arithmetical process; but we
are not required to find these approximate values in
Algebra: for us V^ ^ simply that quantity which when
squared will become 7.
12. A collection of algebraical symbols, that is of letters,
figures, and signs, is called an algebraical expression.
The parts of an algebraical expression which are con
nected by the signs + or — are called the terms.
Thus 2a — Sbx + 5cy' is an algebraical expression con
taining the three terms 2a, — Sbx, and + 5cy\
DEFINITIONS. 5
13. When two terms only differ in their numerical
coefficients they are called like terms. Thus a and 3a are
like terms ; also 6a'6'c and 3a'6*c are like terms.
14. An expression which contains only one term is
called a monomial expression, and expressions which
contain two or more terms are called multinomial expres
sions; expressions which contain two terms, and those
which contain three terms are, however, generally called
binomial and trinomial expressions respectively. Thus
3a6'c is a monomial, a'+ 36' is a binomial, and aa? \hx + c
is a trinomial expression.
15. The sign =, which is read 'equals,' or 'is equal
to,' is placed between two algebraical expressions to denote
that they are equal to one another.
The sign > indicates that the number which precedes
the sign is greater tha/n that which follows it. Thus a>b
means that a is greater than 6.
The sign < indicates that the number which precedes
the sign is less than that which follows it. Thus a<b
means that a is less than b.
The signs i=, :{► and ^^ are used respectively for is not
equal to, is not greater than, and is not less than.
The sign •.' is written for the word because or since.
The sign .'.is written for the word therefore or hence.
16. To denote that an algebraical expression is to be
treated as a whole, it is put between brackets. Thus
(a + 6) c means that b is to be added to a and that the
result is to be multiplied by c ; again (a — 6) (c + d) means
that 6 is to be subtracted from a, and that c{ is to be added
to c, and that then the first result is to be multiplied by
the second ; so also (a + by (c + df means that the cube of
the sum of a and b is to be multiplied by the square of the
sum of c and d.
Brackets are of various shapes : thus, ( ), { }, [J.
Instead of a pair of brackets a line, called a virumlum, is
often drawn over the expression which is to be treated as
DEFINITIONS.
a whole : thus a * 6 — c is equivalent to a — (6 — c), and
J a + 6 is equivalent to V(^ + &)• I^ should be noticed
that where no vinculum or bracket is used, a radical sign
refers only to the number or letter which immediately
follows it : thus *J2a means that the square root of 2 is to
be multiplied by a, whereas J^a means the square root of
2a ; also »Ja •\ x means that a; is to be added to the square
root of a, whereas J a + x means that x is to be added to a
and that the square root of the whole is to be taken.
The line between the numerator and denominator of
a fraction acts as a vinculum, for ^ is the same as
Note. It is important for the student to notice that
every term of an algebraical expression must be added or
subtracted as a whole, as if it were enclosed in brackets.
Thus, in the expression a + bc — die +f, b must be
multiplied by c before addition, and d must be divided by
e before subtraction, just as if the expression were written
a+<6c)(dre)f/.
EXAMPLES.
1. Eind the namerical values of the following expressions in each of
which a=lf &=2, c=3, and (2=4.
(i) 6a+8c362d, (ii) 26a3bc+d,
(iii) ab+Sbc5d, (iv) bccadb,
(v) a+bc+d &nd (vi) bcd+cda+dab+abc,
Ans, 0, 12, 0, 1, 11, 50.
2. If a =3, 2) =1 and c=2, find the namerioal values of
(i) 2a»3&24c8, (ii) 2a^B1^c\
(iii) ^c»i6», (iv) a»+3ac«3a«cc»,
and (v) 2a*b^cBb*c^a2e^i^b,
Ans, 19, 6, 0, 1, 0.
DEFINITIONS. 7
3. Find the values of the following expressions in eaoh of which a = 3^
6=2, c=l andd=0.
(i) (3a + 4d)(26~3c),
(ii) 2a«(6«3c«)d,
(iii) a86«2(a6 + c)«,
(iv) a(6«c») + 6(c»d«) + d(a«c«),
(v) 3(a + 6)2(c + d)~2(6 + c)2(a + d),
, , ., 2a2 26« 2c' 2d2
a^d (VI) __—_ +
6 + c c + a b + d a + b'
Aru, 9, 18, 3, 11, 21, 3.
4. Find the valaes of
Va*^^ J5ab + c, V(6*c«+6*c*) and v^aa+4^+ic2,
when o=6, 6=4, c=3.
^?if . 3, 13, 60, 5.
5. Shew that a^b^ and (a +6) (a 6) are eqnal to one another (i)
when a=2, 6=1; (ii) when a=5, 6=3; and (iii) when 0=12, 6=5.
6. Shew that the expressions
aS68, (a6)(a«+a6+62), (a~6)8+3a6(a6),
and (a + 6)83a6(a + 6)26«
are all equal to one another (i) when a=3, 6=2; (ii) when a=5, 6=1;
and (ii) when a =6, 6=3.
CHAPTER IL
Fundamental Laws.
17. We have said that all concrete quantities must
be measured by the number of times each contains some
unit of its own kind. Now a sum of money may be either
a receipt or a payment^ it may be either a gain or a loss ;
motion along a given straight line may be in either of two
opposite directions ; time may be either before or after some
particular epoch ; and so in very many other cases. Thus
many concrete magnitudes are capable of existing in two
diametrically opposite states : the question then arises
whether these magnitudes can be conveniently distin
guished from one another by special signs.
18. Now whatever kind of quantity we are consider
ing + 4 will stand for what increases that quantity by
4 units, and — 4 will stand for whatever decreases the
quantity by 4 units.
If we are calculating the amount of a man's property
(estimated in pounds), + 4 will stand for whatever increases
his property by £4, that is + 4 stands for £4 that he
possesses, or that is owing to him ; so also — 4 will stand
for whatever decreases his property by £4, that is, — 4 will
stand for £4 that he owes.
If, on the other hand, we are calculating the amount
pf a man's debts, + 4 will stand for whatever increases hig
FUNDAMENTAL LAWS. 9
debts, that is, + 4 will now stand for a debt of £4 ; so also
— 4 will now stand for whatever decreases his debts, that
is, — 4 will stand for £4 that he has, or that is owing to
him.
If we are considering the amount of a man's gains, + 4
will stand for what increases his total gain, that is, + 4
will stand for a gain of 4 ; so also — 4 will stand for what
decreases his total gain, that is, — 4 will stand for a loss of
4. If however we are calculating the amount of a man's
losses, + 4 will stand for a loss of 4, and — 4 will stand for
a gain of 4.
Again, if the magnitude to be increased or diminished
is the distance from any particular place, measured in
any particular direction, + 4 will stand for a distance of
4 units in that direction, and — 4 will stand for a distance
of 4 units in the opposite direction.
19. From the last article it will be seen that it is not
necessary to invent any new signs to distinguish between
quantities of directly opposite kinds, for this can be done
by means of the old signs + and — .
The signs + and — are therefore used in Algebra with
two entirely different meanings. In addition to their
original meaning as signs of the operations of addition and
subtraction respectively, they are also used as marks of
distinction between magnitudes of diametrically opposite
kinds.
The signs + and — are sometimes called signs of
affection when they are thus Used to indicate a quality of
the quantities before whose symbols they are placed.
The sign +, as a sign of affection, is frequently omitted ;
and when neither the + nor the — sign is prefixed to a
term the + sign is to be understood.
20. A quantity to which the sign + is prefixed is
called a positive quantity, and a quantity to which the
sign — is prefixed is called a negative quantity.
The signs + and — are called respectively the positive
md negative signs.
10 FUNDAMENTAL LAWS.
Note. Although there are many signs used in algebra,
the name sign is often used to denote the two signs + and
— exclusively. Thus, when the sign of a quantity is
spoken of, it means the + or — sign which is prefixed to
it; and when we are directed to change the signs of an
expression, it means that we are to change the + or —
before every term into — or + respectively.
21. The magnitude of a quantity considered inde
pendently of its quality, or of its sign, is called its absolute
magnitude. Thus a rise of 4 feet and a fall of 4 feet are
equal in absolute magnitude ; so also + 4 and — 4 are
equal in absolute magnitude, whatever the unit may be.
Addition.
22. The process of finding the result when two or more
quantities are taken together is called addition, and the
result is called the sum.
Since a positive quantity produces an increase, and a
negative quantity produces a decrease, to add a positive
quantity we must add its absolute value, and to add a
negative quantity we must subtract its absolute value.
Thus, when we add f 4 to + 6, we get + 6 + 4 ; and when
we add — 4 to + 10, we get + 10 — 4.
Hence + 6 + (+4) = + 6 + 4,
and + 10 + ( 4) = + 10 ~ 4.
So also, when we add + 6 to + a, we get + a + 6 ; and
when we add — ft to + a, we get + a  ft. Hence
+ a + (+ ft) = + a + i,
and + a + (— ft) = + a — 6.
We therefore have the following nile for the addition
of any term : to add any term affix it to the expression to
which it is to he added, with its sign unchanged.
When numerical values are given to a and to h, the
numerical values of a + 6 and a — 5 can be found ; but
ADDITION OF TERBCS. 11
until it is known what numbers a and b stand for, no
further step can be taken, and the process is considered
to be algebraically complete.
23. When b is greater than a, the arithmetical
operation denoted by a — 6 is impossible. For example, if
a=3 and 6 = 5, a — 6 will be 3 — 5, and we cannot take
5 from 3. But to subtract 5 is the same as to subtract 3
and 2 in succession, so that
35 = 332 = 02 = 2.
We then consider that —2 is 2 which is to be sub
tracted from some other algebraical expression, or that
— 2 is two units of the kind opposite to that represented
by 2 ; and if — 2 is a final result, the latter is the only
view that can be taken.
In some particular cases the quantities under con
sideration may be such that a negative result is without
meaning ; for instance, if we have to find the population
of a town firom certain given conditions ; in this case the
occurrence of a negative result would shew that the given
conditions could not be satisfied, and so also in this case
would the occurrence of a firactional result.
Subtraction.
24. Since subtraction is the inverse operation to that
of addition, to subtract a positive quantity produces a
decrease, and to subtract a negative quantity produces an
increase. Hence to subtract a positive quantity we must
subtract its absolute value, and to subtract a negative
quantity we must add its absolute value. Thus, to
subtract + 4 from + 10, we must decrease the amount by
4 ; we then get +10 — 4.
Also to subtract — 4 from + 6, we must increase the
amount by 4 ; we then get +6 + 4.
Hence +10(+4) = + 104 = + 6,
and + 6(4) = + 6 + 4 = +10.
12 FUNDAMENTAL LAWS.
So also, in all cases
a — (+ 6) = a — &,
and a — (— 6) = a + 6.
We therefore have the following rule for the subtraction
of any term : — to subtract any term affix it to the eocpression
from which it is to be subtracted but vrith its sign changed,
25. We have hitherto supposed that the letters used
to represent quantities were restricted to positive values ;
it would however be very inconvenient to retain this
restriction. In what follows therefore it must always be
understood, unless the contrary is expressly stated, that
each letter may have any positive or negative value.
Since any letter may stand for either a positive or for
a negative quantity, a term preceded by the sign + is not
necessarily a positive quantity in reality ; such terms are
however still called positive terms, because they are so in
appearance; and the terms preceded by the sign — are
similarly called negative terms,
26. On the supposition that b was a positive quantity,
it was proved in Articles 22 and 24, that
a + (+&) = a + 6 (i) ^
a + (6) = a6 (ii) I
a (+6) = a 6 (iii)[ ^^''•
and a — (— 6) = a + 6 (iv) j
We have now to prove that the above laws being true
for all positive values of b must be true also for negative
values.
Let b be negative and equal to — c, where c is any
positive quantity; then
+ 6 = + ( — c) = — from (ii),
and — ft = — ( — c) = Hc from (iv).
Hence, putting — c for + 5, and + c for — ft in (i), (ii),
SUBTRACTION OP TERMS. 18
(iii), (iv), it follows that these relations are true for all
negative values of b, provided
a + {+c)=a + c,
^ ~ (~ c) = a + c,
and a — (+ c) = a — c,
are true for all positive values of c ; and this we know to
be the case.
Hence the laws expressed in (A) are true for all values
of6.
27, Def. The difference between any two quantities
a and b is the result obtained by subtracting the second
from the^r^^.
The algebraical diflference may therefore not be the
same as the arithmetical dififierence, which is the result
obtained by subtracting the less from the greater. The
symbol a ^ 6 is sometimes used to denote the arithmetical
difference of a and 6.
Def. One quantity a is said to be greater than another
quantity b when the algebraical difference a — ft is positive.
From the definition it is easy to see that in the series
1, 2, 3, 4, &c., each number is greater than the one before
it ; and that, in the series — 1, — 2, — 3, — 4, &c., each
number is less than the one before it.
Thus 7, 5, 0, — 5, — 7 are in descending order of
magnitude.
EXAMPLES.
Ex. 1. Find the sum of (i) 5 and  4, (ii)  5 and 4, (iii) 5,3 and
6 and (iv) 3, 4, 6 and 6. Ans, 1, 1, 4, 0.
Ex. 2. Subtract (i) 3 from 4, (ii) 4 from 3, and (iii) a from
 b, Ans.  7» 7, b + a,
Ex. 3. A barometer fell *01 inches one day, it rose *015 inches on
the next day, and fell again *01 inches on the third day. How
much higher was it at &e end than at the beginning?
Ans.  *005 inches.
Ex. 4. A thermometer which stood at 10 degrees centigrade, fell
20 degrees when it was pat into a freezing mixtare: what was
the final reading? Ans. 10.
14 FUNDAMENTAL LAWS.
Ex. 6. Find the value of ab + c and of a+6c, when a=l,
6 = 2andc=3. 4fw.6, 6,
Ex. 6. Find the value of a+bc when
a=l, &= 2, c= 1; also when
a= 2, 6=l, c=3. Alls, 2,4.
Ex.7. Find the value of a (6) + (c) when
a=3, 6=2, c=l. iiiM. 4.
Ex. 8. Find the value of  a+ (  6)  (  c) when
a=  2, 6=  3, c=  6. Ans. 0.
Ex.9. Find the value of (a) + 6(c) when
a=i, 6= 2, c=3. 4rw. 6.
Multiplication.
28. In Arithmetic, multiplication is first defined to be
the taking one number as many times as there are units in
another. Thus, to multiply 5 by 4 is to take as many
fives as there are units in four. As soon, however, as
fractional numbers are considered, it is found necessary to
modify somewhat the meaning of multiplication, for by the
original definition we can only multiply by wlu)le numbers.
The following is therefore taken as the definition of
multiplication : " To multiply one number by a second is to
do to the first what is done to v/nity to obtain the second.**
Thus 4isl + l + l + l;
.. 5 X 4 is 5 + 6 + 5 + 6.
Again, to multiply f by f , we must do to f what is
done to unity to obtain f ; that is, we must divide f into
four equal parts and take three of those parts. Each of
the parts into which f is to be divided will be = — r , and
by taking three of these parts we get = — j. Thus s x j
5x3
"7x4
MULTIPLICATION OF MONOMIAL EXPRESSIONS. 15
Soalso, (5)x4 = (5) + (5) + (6) + (5)
= 20.
With the above definition, multiplication by a negative
quantity presents no difficulty.
For example, to multiply 4 by — 5. Since to subtract
5 by one subtraction is the same as to subtract 5 units
successively,
5 = lllll;
.*. 4x( — 6)=4 — 4 — 4 — 4 — 4
= 20.
Again, to multiply — 5 by — 4. Since
.•.(5)x(4) = (5)(5)(5)(5)
= + 5 + 5 + 5 + 5 [Art.26]
= + 20.
We can proceed in a similar manner for any other
numbers, whether integral or fractional, positive or nega
tive.
Hence we have the following rule :
To find ike product of any two qtuintities, multiply their
absolute values^ and prefijX the sign + if both factors he
positive or both negative, and the sign — if one factor be
positive and the other negative.
Thus we have
( + a)x( + J) = + a6 (i)
(a)x( + 6) = a6 (ii) . .^.
( + a)x(6) = a6 (iii) ^ '"•^ ^'
(a) X (6)= + a6 (iv)
The rule by which the sign of the product is determined
is called the Law of Signs. This law is sometimes
enunciated briefly as follows : Like signs give +, and unlike
signs give — .
16 FtJNDAMENtAL LAWS.
29. The fitctors of a product may be taken in
any order. It is proved in Arithmetic that when one
number, whether integral or fractional, is multiplied by a
second, the result is the same as when the second is multi
plied by the first.
The proof is as follows: when the numbers are integers,
a and b suppose, write down a series of rows of dots,
putting a dots in each row; and take b rows, writing the
dots under one another as in the following scheme :
* * * * * * a in a row
« 4& 4& « « «
« « « 4& 4& «
Ik * * * * *
h rows.
Then the whole number of the dots is a repeated b
times, that is a x 6. Now consider the columns instead
of the rows: there are clearly b dots in each column, and
there are a columns ; thus the whole number of dots is b
repeated a times, that is 6 x a. Hence, when a and b are
integers, ab = ba.
When the numbers are fractions, for example ^ and f ,
we prove as in Art. 28 that ^ x j = = — j . And, by the
, /.px 5x33x6, 5335
above proof for mtegers, = — . = . — = ; hence = x r = 7 x =.
^ ^'7x44x7' 7447
Hence we have ab = ba, for all positive values of a and
6; and the proposition being true for any positive values of
a and ft, it must be true for all values, whether positive or
negative ; for from the preceding Article the absolute value
of the product is independent of the signs, and the sign
of the product is independent of the order of the factors.
Hence for all valves of a and b we have
ab = ba (i).
If in the above scheme we put c in place of each of the
MULTIPLICATION OF MONOMIAL EXPRESSIONS. 17
dots; the whole number of the c's will be ah; also the
number of c's in the first row will be a, and this is repeated
b times. Hence, when a and b are integers, c repeated ab
times gives the same result as c repeated a times and this
repeated b times. So that to multiply by any two whole
numbers in succession gives the same result as to multiply
at once by their product; and the proposition can, as
before, be then proved to be true without restriction to
whole numbers or to positive values. Thus, for all values
of a, b and c, we have
axbx c = ax (be) (ii).
By continued application of (i) and (ii) it is easy to
shew that the factors of a product may be taken in any
order, however many factors there may be. Thus
abc^cab = cba, &c (C).
30. Since the factors of a product may be taken in
any order, we are able to simplify many products. For
example:
3ax4a = 3x4xaxa = 12a*,
( 3a) X ( 46) = + 3a X 46 = 3 X 4 X a X & = 12a6,
(jciby = abxab=:axaxbxb = a^b\
(V2a)' = V2a x ^2a = V2 x V2 x aa = 2a'.
Although the order of the factors in a product is
indifferent, a factor expressed in figures is always put first,
and the letters are usually arranged in alphabetical order.
31. Since a* = aa, and a' = aaa] we have
So also
a^ X a^=iaax aaa = a' = a^'^\
a?xa^ = aaa x axma = a^ = a^"^*.
and a^ xa = aaaa xa = a^ = a*'*'^.
In the above examples we see that the index of the
'product of two powers of the same letter is equal to the sum
of the indices of the factors. We can prove in the following
S.A. 2
18 FUNDAMENTAL LAWS.
manner that this is true whenever the indices are positive
integers :
since by definition
oT = doaa ... to m factors,
and a* = aaa<i ... to n factors;
/. a"* X a" = (aaa... to m factors) x (aaa ... to n factors)
= acm . . . to (tw + n) factors,
= dr^, by definition;
hence a~ y,d!'=^ai^'^ (D).
The law expressed in (D) is called the Index Iiaw.
32. Since ( a) x ( a) = + a* = (+ a) (+ a) [Art. 28],
it follows conversely that the square root of a* is either
+ a or — a: this is written hjc? = ± a, the double sign being
read ' plus or minus/
Thus there are two square roots of any algebraical
quantity, which are equal in absolute magnitude but
opposite in sign.
EXAMPLES.
1. Multiply 2a by 46, a^ by aSand 2a%\)j ^aW,
Am, Sabt a^t 6a*6*.
2. Multiply
2xy2 by ^Sy% Saxhf by 5a^xy^, and da%c^x by 12ab^cxK
Arts, Qxy% 15a^x^y\ S6a^h^(^x*.
3. Multiply
7a*6Sc«by SaSft'c^, and 2ab^x^y^hj ^a^b^x*y^,
Ans. 21a768c9, 8a*6«a:»y8.
4. Find the values of (  a)\ (  a)8, (  a)^ and (  a)».
Ans. a^ a', a*, a".
5. Find the values of (  ab)^, (a^b)^ and (  Zah^c^)K
Ans, a^fta, a^b*, 27a%^cK
6. Shew that the successive powers of a negative quantity are
alternately positive and negative.
7. Find the cubes of 2a«&, dab^c\ and 2a^bxhf'K
Am, Sa^fts,  27a^b^c» and  Qa^b^a^y«.
DIVISION OF MONOMIAL EXPRESSIONS. 19
8. Find the value of (a)ax(6)8, of (  2a6^)« x (  3a»6)», and
of (3a6c)2x(2a»6)>.
Ans. ^a^b^, 216a»6», 72a»6«c«.
9l Rnd the value of dabe2a%€^ + 4c*, when a =2, 6=l, and
c=2.
Afu. 12.
10. Find the value of 2a«6c  36^d + 4c'da  5cPa6, when a=l,
bsz 2, c= 3 and d= 4.
.4n<. 148.
Division.
33. Division is the inverse operation to that of multi
plication; so that to divide a by 6 is to find a quantity c
such that cxb = a.
Since division is the inverse of multiplication and
multiplications can be performed in any order [Art. 29], it
follows that successive divisions can be performed in any
order. Thus aJ6Tc=aic5fc.
It also follows from Art. 29 that to divide by two
quantities in succession gives the same result as to divide
at once by their product. Thus ai6rO = ar (6c), which
is usually written a r be.
Not only may a succession of divisions be performed
in any order, but divisions and multiplications together
may be performed in any order. For example
axbic = a'^cxb.
For a = a7cxc;
.*. ax6 = arcxcx6
= a~cx6xc; [by Art. 29]
therefore, dividing each by c, we have
ax6Tc = arcx6.
Hence we get the same result whether we divide the
product of a and b by c, or divide a by c and then multiply
by 5, or divide 6 by c and then multiply by a,
34. The operation of division is often indicated by
placing the dividend over the divisor with a line between
2—2
20 FUNDAMENTAL LAWS.
them: thus t means arb. Sometimes a/b is written for
7 . When aTb is written in the fractional form 7 , a is
called the numerator, and b the denominator.
Since  = 1 7 c,
c
xc=l^cxc = l.
c
Also axxc = ax(xc)=axl=a.
Therefore, dividing by c,
1
ax ^aic,
c
so that to divide by any quantity c is the same as to
multiply by the quantity  .
Hence axbrc^arcxb,
can be written,
ax6x = axx6,
c c
in which form it is seen to be included in Art. 29 (0),
35. Since a? xci^ — a", and a^ xa^= a*®; we have con
versely a* "T a' = a', and a*® h a' = a'.
And, in general, when m and n are any positive integers
and m>n,vre have
for by Art. 31
a"*^a* = a'^.
a"' " X a* = a*.
Hence if one power of any quantity be divided by
a lower power of the same quantity, the index of the
Quotient is equal to the difi^erence of the indices of the
ividend and the divisor.
DIVISION OF MONOMIAL EXPRESSIONS. 21
Hence a^V  a*6 = a* x 6' r a' ^ 6 = a* 7 a* x 6" 4 6 = a%
and a'6VTa'6V = aW.
36. We have proved in Art. 28 that
a X (— 6) = — a6 ;
/. (— ab) T (— 6) = a, and (— oft) t a = — 6 ;
we have also proved that
(a)(6) = + a6 = (+a)(+6);
.'. (+ 06) r ( a) =  fc, and (+06) 7 (+ a) = + 6.
Hence if the signs of the dividend and divisor are
alike, the sign of the quotient is + ; and if the signs of
the dividend and divisor are unlike, the sign of the quo
tient is — ; we therefore have the same Iiaw of Signs in
division as in multiplication.
Thus a'6«^ai» = a''fe^
and  2a^W ^  3a*6c' =  ac\
EXAMPLES.
1. Divide 10a by  2a, Sa^ft* by  2ah\ and  7a56»c* by  3a«6V.
2. Divide 2a^6^c' by 4a'6c'^, 6a^y* by Zxhf^ and ^o^h^xfy^ by
2a6*iBay6,
1 /»
8. Multiply  2a'6c' by  ^aJiPc^ and divide the result by 8a'6'c«.
Am, rdb^e,
4
37. The fundamental laws of Algebra, so far as
monomial expressions are concerned, are those which were
24 FUNDAMENTAL LAWS.
42. If c be any positive integer, a and b having any
values whatever, then
(a + 6)c = (a + 6) + (a+6) + (a + J)+ ... repeated c times
= a + 6 + a + 6 + a + 6^+ ... [Art. 40]
= a + a + a + . . . repeated c times
+ b + b + b + ... repeated c times
= ac + be.
Hence, when c is a positive integer, we have
(a'\b)c=ac + bc (F).
Since division is the inverse of multiplication, it follows
that when d is any positive integer
(a + b) z d = a i d + b 7 d.
And hence
(a+b) X c 7 d = {(a + b) X c] z d
= (ac + bc)^d = ac7d+bcid,
that is (a + 6)Xj = ax^ + feXj.
^ d d d
Thus the law expressed in (F) is true for all positive
values of c ; and being true for any positive value of c, it
must also be true for any negative value. For, if
(a + 6) c = ac + 6c,
then (a + 6) (— c) =  (a 4 6) c = — oc  6c
= a (— c) 4 6 (— c).
Hence for all values of a, b and c we have
{a^b)o^ac\bc (F).
Thus the product of the sum of any two algebraical
quantities by a third is the sum of the products obtained
by multiplying the quantities separately by the third.
The above is generally called the Distributive Law.
MULTINOMIAL EXPRESSIONS. 25
43. Since (a + 6) r c = (a + 6) x 
c
c c
we see that the quotient obtained by dividing the sum of
any two algebraical quantities by a third is the sum of the
quotients obtained by dividing the quantities separately by
the third.
44. From Art. 40 it follows that
a + 6 + c + d + 6 4 . . . = (a + 6) + c + (d + e) + . . .
= a + (6 + c + d) + e+ ... = &c.,
so that the terms of an expression may be grouped in any
manner.
Again, from Art. 29, it follows that
abcde . . . = a {be) (de) . . . = a {bed} e . . . = &c.,
so that the factors of a product may be grouped in any
manner.
These two results are called the Associative Iiaw.
45. We have now considered all the fundamental laws
of Algebra, and in the succeeding chapters we have only to
develope the consequences of these laws.
CHAPTER III.
Addition. Subtraction. Brackets.
Addition.
46. We have already seen that any term is added by
writing it down, with its sign unchanged, after the expres
sion to which it is to be added ; and we have also seen
that to add any expression as a whole gives the same
result as to add all its terms in succession. We therefore
have the following rule : — to add two or more algebraical
encpressionSy write down all the terms in succession with
their signs unchanged.
Thus the sum of a — 26 + 3c and — 4c? — 5e + 6/* is
a26 + 3c4d 5e + 6/.
47. If some of the terms which are to be added are
'like' terms, the result can, and must, be simplified before
the process is considered to be complete.
Now two *like' terms which have the same sign are
added by taking the arithmetical sum of their numerical
coefl&cients with the common sign, and affixing the com
mon letters.
For example, to add 2a and 5a in succession gives the same result,
whatever a may be, as to add 7a; that is, +2o+5o= +7a. Also, to
subtract 2a and 5a in succession gives the same result as to subtract
7a; that is,  2a  5a = 7a.
ADDITION. SUBTRACTION. 27
Also two 'like* terms whose signs are diflferent are
added by taking .the arithmetical difference of their
numericsd coefficients with the sign of the greater, and
affixing the common letters.
For example, +6a3a=+2a+3a3a=+2a,
also +3a6a= +3a3a2a= 2a.
Thus, when there are several 'like' terms some of
which are positive and some negative, they can all be
reduced to one term.
Ex. 1. Add 2a + 5b to a^b.
TheBomis a^b+2a+5h
=a+2a66 + 66
=8a6.
Ex.2. Add Sa^5ab + W, 4a«2a6 + 36«,
and 2a'+6o686«.
Thesmni8 3a«6a6 + 7&'4a'2a6 + 3&2+2a«+6a6862.
The terms 3a',  4a', and + 2a' can be combined mentally; and we
have a'. Similarly we have  2db and + 2&'.
Thus the required smn is a'  2ab + 2&'.
The beginner will find it desirable to put like terms
under one another.
Subtraction.
48. We have already seen that any term may be sub
tracted by writing it down, with its sigii changed/after the
expression from which it is to be subtracted ; and we have
also seen that to subtract any expression as a whole gives
the same result as to subtract its terms in succession. We
therefore have the following rule : To subtract any alge
braical expression, write down its terms in succession with
all the signs changed.
Thus, if a — 26 + 3c be subtracted from 2a — 36 — 4c,
the result will be 2a  36 — 4c — a + 26 — 3c = a — 6 — 7c.
49. The expression which is to be subtracted is some
times placed under that from which it is to be taken, 'like'
terms being for convenience placed under one another;
28 BRACKETS.
and the signs of the lower line are changed mentally before
combining the 'like' terms.
Thus the previous example would be written down as under:
2a364c
a26 + 3c
a blc
As another example, if we have to subtract SabSuc+c^ from
o^  5a^ + 2ac  2b\ the process is written
a26a6+2ac262
Sab  5ac + c*
a2_8a6 + 7ac262ca"
Brackets.
50. To indicate that an expression is to be added as a
whole, it is put in a bracket with the + sign prefixed.
But, as we have seen in Art. 46, to add any algebraical
expression we have only to write down the terms in suc
cession with their signs unchanged.
Hence, when a bracket is preceded by a + sign, the
bracket may be omitted.
Thus + (2a  66 + 7c) = + 2a  56 + 7c.
Hence also, any number of terms of an expression may
be enclosed in brackets with the sign + placed before each
bracket. Thus
3a26 + 4cd + c/=3a26 + (4cd + e/)
= 3a + (26 + 4c)  d + (c /).
When the sign of the first term in a bracket is + it
is generally omitted for shortness, as in the preceding
example.
51. To indicate that an expression is to be subtracted
as a whole, it is put in a bracket with the — sign prefixed.
But, as we have seen in Art. 48, to subtract any alge
braical expression we have only to write down the terms
in succession with all their signs changed.
BRACKETS. 29
Hence, when a bracket is preceded by a — sign, the
bracket may be omitted, provided that the signs of all the
terms within the bracket are changed. Thus
a(26c4c?) = a26 + cd
Hence also, any number of terms of an expression may
be enclosed in a bracket with the sign — prefixed, provided
that the signs of all the terms which are placed in the
bra.cket are changed. Thus
a — 26 + 3c  d = a  (26  3c + d) = a  2 J  ( 3c + d).
52. Sometimes brackets are put within brackets : in
this case the different brackets must be of different shapes
to prevent confusion.
Thus a — [26 — {3c — (2d — e)}] ; which means that we
are to subtract from 26 the whole quantity within the
bracket marked { }, and then subtract the result from a ;
and, to find the quantity within the bracket marked { },
we must subtract e from 2d, and then subtract the result
from 3c.
When there are several pairs of brackets they may be
removed one at a time by the rules of Arts. 50 and 51.
Thus a  [6 4 {c  (d  e)}]
= a[6 + {cd4c}]
= a— [6 + c— d + e]
= flt — J — c + d — c.
EXAMPLES I.
1. Add Sx  5y, 505 — 2y and 7y — 4aj.
2. Add 3a5  5y + 2«, 5a5 — 7y — 5« and 6y — «  lOos.
3. Add ^a^b + 1<;, ^6  ^c + ^a and Jc  ^a + ^b.
4u Add a'  a* + a, a*  a + 1 and a^ — a^— 1.
6. Add as"  6xy — 7^ and St/* + 4aw/  as*.
30 EXAMPLES.
6. Add vr?  3mn + 2n', 3n^  m^ and 5mn  Sn^ + 2m'.
7. Add 3a"  2ac  2a6, 26" + 36o + 3a6 and c*  2a(5  2i(j.
8. Add f a»6  5a6* + 76^ 2a«  a'6 + 5a6" and 36»  2a^
9. Subtract 3a — 46 + 2c from a + b —2c,
10. Subtract « + 9 ^  o <^ ^J'om c  ^ a  ^ 6.
11. Subtract 3a:"  4aj + 2 from 4iB"  5a;  7.
12. Subtract 5a*  3a«6 + 4a'i' from 56*  3a6« + 4a"6*.
13. What is the difference between  305"  5xy + 4^* and
5af + 2xy3f'i
14. What must be added to 26c — 3ca — 4a5 in order that
the sum may be hc + ca'i
15. What must be added to 3a' 26' + 30* in order that
the sum may be 6c + ca + a6?
16. Simplify 3x  {2y + {5x  dx + 3/)}.
17. Simplify x[3y + {3z  x^^} + 2x].
18. Simplify y2x{zx — yx + z}.
19. Simplify a[a6 — {a — 6 + c — a — 6 + c d}],
20. Simplify 2x  [3x  9y  {2x  31/  (a; + 6y)}].
21. Simplify a  [3a + c  {4a  (36  c) + 36}  2a],
22. Subtract x  (Zy  z) from ^ — {2aj  » — y}.
23. Subtract 2m — (3m  2ti — m) from 2w  (37i  2m — n).
24. Find the value of
{a{b c)Y + {6  (c  a)}' + {c  (a  6)}' when
a = l, 6 = 2, c=3.
25. Eind the yalae of
{a'  (6  cY}  {6'  (c  a)'}  {c'  (a  6)'} when
a=l, 6=2, c= 3.
CHAPTER IV.
Multiplication.
53. Product of monomial expresBioiiB. The
multiplication of monomial expressions was considered in
Chapter II., and the results arrived at were :
(i) The factors of a product may be taken in any
order.
(ii) The sign of the product of two quantities is +
when both the factors are positive or both negative ; and
the sign of the product is — when one factor is positive
and the other negative.
(iii) The index of the product of any two powers of
the same quantity is the sum of the indices of the factors.
From (i), (ii) and (iii) we can find the continued
product of any number of monomial expressions.
Thus (  2a«6c») x (  Sa^b^c) = + 2a^bc^ x Sa^li^c, from (ii),
=2x8xa2.a'.6.6*.c8.c, from (i),
=6a'6«c*, from (iii).
Again, ( 8a»6) ( 6a58) ( 7o*63) = {+ 3a«6 . 6a68 (( 7a«62)
= 8.6.7.a3.a.a*.6.68.6»=106a76«.
54. Prodnot of a multinomial expressioii and a
monomial. It was proved in Art. 42 that the product
of the sum of any two algebraical quantities by a third is
equal to the sum of the products obtained by multiplying
the two quantities separately by the third.
32 MULTIPLICATION.
Thus (X'\y) z = x£r + yz (i).
Since (i) is true for all values of oc, y and z, it will be
true when we put (a + 6) in place of x ; hence
{{a + b) ^y] z = {a + b) z {■ yz
= az + bz + yz,
/. (a + b + y)z = az\'bz\ yz.
And similarly
(a + 6 + c + d + ,..) ^ = asr + 6^ + cs: + d^ + ...,
however many terms there may be in the expression
a + 6 + c+d4...
Thus the product of any multinomial expression by a
monomial is the smti of the products obtained by multiplying
the separate terms of the multinomial eoopression by the
monomial,
55. Product of two multinomial expreBsions.
We now consider the most general case of multiplication,
namely the multiplication of any two multinomial ex
pressions.
We have to find
(a\b\c + ,,,)x(x + y + z+ .,.);
and, from Art. 38, this includes all possible cases.
Put M for x + y + Z+ ...; then, by the last article,
we have
(a\b + o+...)M=aM\bM\cM{' ,..
=^Ma + Mb'\Mc'\ ...
=^(x\y + z+ ..,)a\'(x\'y\z+ ,,.)b
'\'{x + y + z+..,)o+ ...
=^ax + ay + az+..,\bx\'by\'bz+,,. + cx + cy + cz+.,.
Hence (a + 64c+ ...) {x + y+ z + ,,,)
=:ax+ay + az+ ,,, +bx^by ^bz^ ,,, hcx + cy + cz^r '..
MULTIPLICATION. 33
Thus, the product of any two algebraical expressions is
eqtml to the sum of the products obtained by mvltvplying
every term of the one by every term of the other.
For example
(a + b) {c + d) = ac '\ ad ■{' be '\bd ;
also
(3a + 56) (2a + 36)
= (3a) (2a) + (3a) (36) + (56) (2a) + (56) (36)
= 6a* 4 9a6 + 10a6 + 156' = 6a» + 19a6 4 156'.
Again, to find (a ^ 6) (c — d), we first write this in the
form [a +( 6)} [c +(— d!)}, and we then have for the product
oc + a ( d) + ( 6) c + ( 6) (~ d)
=^ac — ad — 6c + bd.
In the rule given above for the multiplication of two
algebraical expressions it must be borne in mind that the
terms include the prefixed signs.
56. The following are important examples : —
I. (a + 6)' = (a + 6) (a + 6) = aa + a6 + 6a + 66 ;
.. (a + 6)' = a» + 2a6 + 6».
Thus, the square of the sum of any two quantities is
equal to the sum of their squares plus twice their product
II. (a  6)' = (a  6) (a  6) = aa + a ( 6) + ( 6) a
+ (6)(6) = a«a6a6 + 6';
.. (a6)* = a*2a6 + 6'.
Thus, the square of the difference of any two quantities is
equal to the sum of their squares minus twice their product.
III. (a + 6)(a6) = aa + a(~6) + 6a + 6(6)
= a' — a6 + a6  6' ;
.. (a + 6)(a6) = a»6'.
Thus, the product of the sum and difference of any two
quantities is equal to the difference of their squxares.
S. A. 3
34 MtTLTlPLlOATlON.
57. It is usual to exhibit the process of multiplication
in the following convenient form :
a» + 2aft  V
a* + 2a'6  a«fe«
 2a'h  4a'6« 4 2a6'
a*6' + 2a6^6*
a*  4a*6* + 4a6»  h\
The multiplier is placed under the multiplicand and a
line is drawn. The successive terms of the multiplicand,
namely a', + 2a6, and — 6', are multiplied by a^ the first
term on the left of the multiplier, and the products a*,
+ 2a'6 and — a*6' which are thus obtained are put in a
horizontal row. The terms of the multiplicand are then
multiplied by — 2a6, the second term of the multiplier,
and the products thus obtained are put in another hori
zontal row, the terms being so placed that 'like' terms
are under one another. And similarly for all the other
terms of the multiplier. The final result is then obtained
by adding the rows of partial products ; and this final
sum can be readily written down, since the difierent sets
of * like' terms are in vertical columns.
The following are examples of multiplication arranged
as above described :
a+h a — h a — h
a^ + ab a^ + ab a^ + a^h + ah'
+ ab + b* aJblfl d^baV^lfl
a« + 2ab + b^ a^ ^^ a^ Tp
a\b\c 8x«a;y + 2y2
a+b + c %sx^+xy2y^
a^ + ab + ac 9a5*  3a:«y + 6j; V
+ ab +6«+ be +8aB>y xhf^^^xy^
+ac + &C + C* 6a;V + 2a;y»~4y*
68. If in an expression consisting of several terms
which contain different powers of the same letter, the
MULTIPLICATION. 85
tenn which contains the highest power of that letter be put
first on the left, the term which contains the next highest
power be put next, and so on ; the terms, if any, which
do not contain the letter being put last ; then the whole
expression is said to be arranged according to descending
powers of that letter. Thus the expression
is arranged according to descending powers of a. In like
manner we say that the expression is arranged according
to ascending powers of b.
69. Although it is not necessary to arrange the terms
either of the multiplicand or of the multiplier in any
particular order, it will be found convenient to arrange
both expressions according to descending or both according
to ascending powers of the same letter: some trouble in
the arrangement of the different sets of 'like' terms in
vertical columns will thus be avoided.
60. DeflnitionB. A term which is the product of n
letters is said to be of n dimensions, or of the nth degree.
Thus Sabc is of three dimensions, or of the third degree ;
and 5a*b% that is 5aaa&frc, is of six dimensions, or of the
sixth degree. Thus the degree of a term is found by taking
the sum of the indices of its factors.
The degree of an expression is the degree of that term
of it which is ofhighest dimensions. ^
In estimating the degree of a term, or of an expression,
we sometimes take into account only a particular letter, or
particular letters : thus aa? + 6a; + o is of the second degree
in X, and is often called a quadratic expression in x ; also
aa?y + bxyhca^ is of the third degree in x and y, and is
often called a cviic expression in x and y. An expression,
or a term, which does not contain x is said to be of no
degree in x, or to be independent of x.
When all the terms of an expression are of the same
dimensions, the expression is said to be homogeneous.
Thus a' + 3a'6 — 56' is a homogeneous expression, every
3—2
86 MULTIPLICATION.
term being of the third degree ; also cux? + hxy + cy^ is a
homogeneous expression of the second degree in x and y,
61. Product of homogeneouB expresidoiiB. The
Eroduct of any two homogeneous expressions must be
omogeneous ; for the different terms of the product are
obtained by multiplying any term of the multiplicand by
any term of the multiplier, and the number of dimensions
in the product of any two monomials is clearly the sum
of the number of dimensions in the separate quantities;
hence if all the terms of the multiplicand are of the
same degree, as also all the terms of the multiplier, it
follows that all the terms of the product are of the same
degree ; and it also follows that the degree of the product
is the swm of the degrees of the factors.
The fact that two expressions which are to be multi
plied are homogeneous should in all cases be noticed ; and
if the product obtained is not homogeneous, it is clear
that there is an error.
62. It is of importance to notice that, in the product
of two algebraical expressions, the term which is of highest
degree in a particular letter is the product of the terms
in the factors which are of highest degree in that letter,
and the term of lowest degree is the product of the terms
which are of lowest degree in the factors: thus there is only
one term of highest degree and one term of lowest degree.
63. Detached Coefficients. When two expressions
are both arranged according to descending, or to ascending,
powers of some letter, much of the labour of multiplication
can be saved by writing down the coefficients only.
Thus, to multiply 3a:* — « + 2 by 3a^ + 2a?  2, we write
31 + 2
3 + 22
93 + 6
62 + 4
6+24
9+32+64
MULTIPLICATION. 37
The highest power of x in the product is clearly w\
and the rest follow in order. Hence the required product
is 9a:* + 3a;» 2«'+ 6a?  4.
When some of the powers are absent their places must
be supplied by O's,
Thus, to multiply a?*  2a;* + a?  3 by a^ + a;'  a?  3,
we write
1+02+13
l.fl4,0,l,3
1+02+13
1+02+13
10+21+3
30+63+9
1+12251+5+0+9
Hence the product is
a? + a?'  2a;*  2a;P  6a?*  a;* + 5a;^ + 9.
This is generally called the method of detached
coefficients.
64. We now return to the three important cases of
multiplication considered in Art. 56, namely,
(a + 6)» = a' + 2a6 + 6' (i),
(a6)« = a*2a6 + 6' (ii),
(a + 6)(a6)=a'6» (iii).
A general result expressed by means of symbols is
called B, formula.
Since the laws from which the above formulae were
deduced were proved to be true for all algebraical
quantities whatever, we may substitute for a and for b
any other algebraical quantities, or algebraical expressions,
and the results will stUl hold good.
88 MULTIPLICATION.
We give some examples of results obtained by substi
tution.
Put — 6 in the place of 6 in (i); we then have
{«+ (6)}' = a« + 2a ( b) + (6)«,
that is (a  by = a'  2ah+b\
Thus (ii) is seen to be really included in (i).
Put \/2 in the place of 6 in (iii) ; we then have
(a + V2) (a V2) =a' (V2)' = a'  2.
[We here, however, assume that all the fundamental
laws are true for surds: this will be considered in a
subsequent chapter.]
Put 6 + c in the place of b in (i); we then have
{a4(6 + c)}' = a'+2a(6 + c) + (6 + c)';
/. (a + 6 + c)' = a' + 2a6 + 2ac + 6' + 26c + c' (iv).
Now put — c for c in (iv), and we have
{a + 6 + (c)}' = a* + 2a6+2a(c) + 6* + 26(c) + (c)»;
.. (a f 6c)'= a* + 2a6  2ac + fe'  2bo+c\
Put 6 + c in the place of 6 in (iii); we then have
{a + (6+c)}{a(6 + c)}=a«(6 + c)« = a'(6'+26c + c«);
.'. (a + 6 + c) (a  6  c) = a'  6*  26c ~ c\
The following are additional examples of products
which can be written down at once.
(a«+26«)(o«26«)=(aV(26«)«=a446*.
(a» + V86«) (a3  ^/86>) = (a^)*  (^/36a)3= a<  86*.
(a6+c)(a+6c)={a(6c)}{a+(6c)}=o>(6c)*.
(a>+a6+6«)(a>a6+6») = {(a«+6«)+a6}{(a«+6«)a6}
= (a«+63)«(a6)«=a*+o«6«+6*.
(a»+«»+«+l)(K»a?+afl)={(«?+«) + («*+l)}{(a;»+a;)(««+l)}
MULTIPLICATION. 39
65. Square of a multinoinial expreiston. We
have found in the preceding Article, and also by direct
multiplication in Art. 57, the square of the sum of three
algebraical quantities; and the square of the sum of
more than three quantities can be obtained by the same
methods. The square of any multinomial expression can
however best be found in the following manner.
We have to find
(a + 6 + c + d+ ...)(a + 6 + c + d+ ...)•
Now we know that the product of any two algebraical
expressions is equal to the sum of the partial products
obtained by multiplying every term of one expression by
every term of the other. If we multiply the term a of
the multiplicand by the term a of the multiplier, we
obtain the term a' of the product: we similarly obtain
the terms 6*, c*, &c. We can multiply any term, say 6,
of the multiplicand by any different term, say d, of the
multiplier; and we thus obtain the term bd of the
product. But we also obtain the term bd by multiplying
the term d of the multiplicand by the term b of the
multiplier, and the term bd can be obtained in no other
way, so that every such term as 6d, in which the letters
are different, occurs twice in the product. The required
product is therefore the sum of the squares of all the
quantities a, 6, c, &c. together with twice the product
of every pair.
Thus, the square of the sum of any number ofalffehrai
cal quantities is equal to the sum of their squares together
with twice the product of every pair.
For example, to find {a+b + c)K
The sqnareB of the separate terms are a^, l\ (?,
The products of the different pairs of terms are a&, ac and he.
Hence (a+6 + c)2=a?+62 + ca+2a6+2ac+26c.
Similarly,
(a+268c)2=a9+(26)a+(3c)a+2a(26) + 2a(3c) + 2(26)(8c)
40 MULTIPLICATION.
And
(a_6 + c:d)«=a2+(6)2 + c«+(d)2+2a(&) + 2ac + 2a(d)
 2ad  2bc + 2bd  2cd,
After a little practice the intermediate steps should be omitted
and the final result written down at once. To ensure taking twice
the product of every pair it is best to take twice the product of each
term and of every term which follows it.
66. Continued Products. The continued product
of several algebraical expressions is obtained by finding
the product of any two of the expressions, and then
multiplying this product by a third expression, and so on.
For example, to find (x \a){x\ b) {x + c), we have
x + a
x + b
+ hx+<ib
x^+(a+b)x+db
gg+c
a:^\{a+b)x* + dbx
\ cs^ + {a+ b)cx \ ahc
xr^ + {a + b+c)x*+((ib + acbc)x + abe
In the above all the terms which contain the same powers of x are
collected together : it is frequently necessary to arrange expressions
in this way.
Again, to find (ob? + a^)^ (x + af {x  a)K
The factors can be taken in any order ; hence the required product
«[(a;a)(x + a)({B» + a«)P=[(a:?a2)(a:a+a«)]«=(a;*a*)2=a;82a^+a».
67. We have proved in Art. 56 that the product of
any two multinomial expressions is the sum of all the
partial products obtained by multiplying any term of one
expression by any term of the other.
To find the continued product of three expressions we
must therefore multiply each of the terms in the product
of the first two expressions by each of the terms in the
third; hence the continued product is the sum of all the
partial products which can be obtained by multiplying
together any term of the first, any term of the second, and
any term of the third.
MULTIPUCATION. 41
And similarly, the continued product of any number of
expressions is the sum of all the partial products which
can be obtained by multiplying together any term of the
first, any term of the second, any term of the third, &c.
For example, if we take a letter from each of the three
factors of
(a + 6) (a + b) {a + 6),
and multiply the three together, we shall obtain a term
of the continued product; and if we do this in every
possible way we shall obtain all the terms of the continued
product.
Now we can take a every time, and we can do this in
only one way; hence a' is a term of the continued
product
We can take a twice and b once, and this can be 5one
in three ways, for the b can be taken from either of the
three binomial factors; hence we have 3a'6.
We can take a once and b twice, and we can do this
also in three ways; hence we have Soft'.
Finally, we can take b every time, and this can be done
in only one way; hence we have 6'.
Thus the continued product is
a' + Sa*b + Sab* + b\
that is (a + 6)' = a' + Sa*b + Sab* + b\
The continued product (x + a) {x + b) {x + c) can simi
larly be written down at once.
For we can take x every time: we thus get a?'.
We can take two x'b and either a or 6 or c : we thus
have a^a, x*b and a^c.
We can take one x and any two of a, 6, c : we thus
have xab, xac, and xbc.
Finally, if we take no a?'s, we have the term oho.
42 MULTIPLICATION.
Thus, arranging the result according to powers oia, we
have
{x + a) {x + b) {x + c) == a? + a^ (a + b + c) + X (ab { ac {■ be)
+ ahc.
68. Powers of a binomial. We have already found
the square and the cube of a binomial expression; and
higher powers can be obtained in succession by actual
multiplication. The method of detached coeflScients should
be used to shorten the work.
The following should be remembered:
(a + 6)' = a'* + 2a6 + 6*,
(a + 6)" = a» + 3a*6 + 3a6'* + 6',
and . (a + by = a* + 4a'6 + 6a*¥ + 4a6» + b\
To find any power, higher than the fourth, of a binomial
expression a formula called the Binomial Theorem should
be employed: this theorem will be considered in a subse
quent chapter.
EXAMPLES IL
1. Multiply 2x — a by as — 2a.
2. Multiply 3a:  J by ^x 3.
3. Multiply x^ + x+l bya;l,
4. Multiply oc^^xy + y* by x + y,
5. Multiply l+aj+ aj* + 05* by xl.
6. Multiply aj* + aj'y + ajy + a;y' + y* by yx,
7. Multiply a5*a; + 2 by iB' + a:2.
8. Multiply 1 + oa; + a'x' by 1 — oaj + a V.
9. Multiply a* + a* + 1 by a;*  a* + L
EXAMPLEa 43
10. Multiply 3aj'ir2^ + 2y* by 3y"ajy + 2a:'.
11. Multiply aj"  5a5* + 1 by 2aj'+5aj + l.
12. Multiply 2a:^  5a5*y + y* by ^ + 5a^ + 2a^.
13. Multiply 3a» 2a»6 + 3a6"  36» by 2a«+ 5a«64a5»+ft».
14. Multiply 2aV  3aVy' + 5/ by aV + 4aajy*  2y^
15. Multiply 2a  3a« + Sa""  7a* by 1  2a* + 6a*.
16. Multiply a' — a6  oc + 6'  6c + c* by a + bkc,
17. Multiply a^ + f/' + a^yz — zx^xy by a5 + y+».
18. Multiply 4a"+ 96" + c" + 36c + 2ca  6a6 by 2a + 36  c.
19. Multiply together a* + 1, as" + 1 and a?'  1.
20. Multiply together x*+l6y\ a5*+ 4^, x + 2y and 05  2y.
21. Multiply together (x — y)', (as + y)' and («■ + y*)'.
22. Multiply together (x' + 1)», (aj + 1)* and (x  1)».
23. Multiply together a'— 05+1, 05*+05+l and a;* 05* + 1.
24. Multiply together a*2a6 + 46*, a* + 2a6 + 46* and
a*4aW+166\
25. Find the squares of (i) a + 26 3c, (ii) a" — a6 + 6',
(iii) 6c + oa + a6, (iv) 1  2aj + 3aj', and (v) a^ + a^ + a: + 1.
26. Find the cubes of (i) a + 6 + c, (2) 2a  36  2c and
(iii) 1+05 + 05*.
27. Simplify
{x + y + zy~{x + y + zY + (xi/ + zy^(x + yzy.
28. Shew that
(aj + y)(aj + 2;)a;"=(2/ + 2;)(y + aj)y» = (« + a;)(2; + y)»".
29. Shew that
(y + «)*+(» + aj)*+(a; + y)*aj*y'2;*=(a; + y + 2;)'.
44 EXAMPLES.
30. Simplify {a: (05 + a)  a (a? — a)} {as (a? — a)  a (05 + a)},
31. Shew that
(y2;)"+ (««)"+ (ajy)» = 3 (y2;)(»aj)(ajy).
32. Shew that a* + 6' = (a + by  3a6 (a + 6), and that
a*+ ft*= (a + by iab{a + by + 2a*ft".
33. Shew that (aj* + ajy + y)"  4a5y (aj* + y*) = (aj*  a^ + y")*.
34. Shew that
(y + «)' + (« + aj)' + (aj + y)' + 2 (aj + y) (a; + ») + 2 (y + ») (y + a?)
+ 2 (» + aj) (» + y) = 4 (aj + y + »)'.
35. Shew that (a« + b') (c« + cf •) = (oc + bdy +{ad bey.
36. Shew that, if x = a + d, y = b + d, and 2? = c + c? ; then
will a' + y* + «'y««a;a^ = a' + 6' + c"6cca — 06.
37. Shew that, ifaj = 6 + c, y = c + a, and » = a + ft; then
will a5' + y' + «* — y2;2fa; — ajy = a*+ft'+c*— ftc — caa6.
38. Shew that 2(a6)(ac) + 2(6c)(6a)+2(ca)(c6)
= (6c)« + (ca)«+(a6)».
39. Shew that (a* + y* + »^ (a* + 6* + c*)  (oa: + 6y + (»)•
= {bz — cy)" + (<»? — azy + (ay — ftx)*.
40. Shew that, if x = a^—bc, y = b* — ca, z = c^^ab; then
will ax + by + cz = {x + y + z)(a + b+c)f
and be (pc^ — yz) = ca (^ — zx) = ab {9^ — xy).
41. Find the value of
{x  a)" + (a;  &)' + (a;  c)"  3 (a:  a) (a;  ft) (a:  c)
when 3a3 = a + & + c.
42. Shew that (a» + 6" + c")*
= (6" + cy + (a6 + ac)" + (a6ac)* + a*
= (6c + ca + a5)" + (a"  6c)« + (&»  ca)» + (c« ^ a6)«.
43. Shew that (a" + ajy + y") (a" + 06 + 6')
« (oaj  byy + (cMC  6y) (ay + fta: + 6y) + (ay + fta; + 6y)°,
EXAMPLES. 45
44 . Shew that 1 + a' + 6' + c' + 6 V + c V + aV + a«& V
= (1 — 6c  ca  aft)' + (a + 6 + c  oftc)*.
45. Shew that
(a* + 6» + c« + c^')^ = (a' + h^ c' <;«)»+ 4(ac + ftf^)»+ ^{ad hc)\
46. Shew that
(i) (a + 2)»4(a+l)«+6a»4(al)» + (a2)«=0.
(ii) (a+2)(6 + 2)4(a + l)(6 + l) + 6aft
4(al)(6l) + (a2)(62) = 0.
47. Shew that
(i) (a + 2)» 4 (a + 1)«+ 6a^ 4(a 1)«+ (a  2)' = 0.
(ii) (a> 2)(6 + 2)(c + 2) 4 (a + 1)(6 + l)(c + 1) + 6a6c
4(al)(6l)(cl) + (a2)(62)(c2) = 0.
48. Shew that
(a + 6 + c)' + (6 + c  a) (c + a  6) (a + 6  c)
= 4a" (ft + c) + 46' (c + a) + 4c' (a + 6) + 4a6c.
49. Shew that
+ z{~ X'¥y+z)(x — y + z) + ( x + y + z)(x y + z) (x + y  z)
= 4:xyz,
60. Multiply
a' + b'^c'^d^ — bcca — ahad — bd — cd by a + 6 + c + cf.
61. Shew that
((»'+»+ l)(a:»^a;+ 1) (aj*(B'+ 1) (x«a*+ 1). . .(a;«"iB'"""+ 1)
= ar +05 +1.
CHAPTER V.
Division.
69. Division by a monomial expression. We
have already considered the division of one monomial
expression by another. We have also seen (Art. 43) that
the quotient obtained by dividing the sum of two alge
braical quantities by a third is the sum of the quotients
obtained by dividing the quantities separately by the
third; and we can shew by the method of Art. 54 that
when any multinomial expression is divided by a monomial
the quotient is the sum of the quotients obtained by
dividing the separate terms of the multinomial expression
by that monomial.
Thus {a^x — 3cw?) = aa? = d?x f oa; — 3aa? ^ oa: = a — 8.
And (12aj'6aa;»2a*a?)H3a?=12a;'^aB Soa^'^ai?
 2a'a; r 3a: = 4c*  foa;  fa*.
70. Division by a multinomial expression. We
have now to consider the most general case of division,
namely the division of one multinomial expression by
another.
Since division is the inverse of multiplication, what
we have to do is to find the algebraical expression which,
when multiplied by the divisor, will produce the dividend.
Both dividend and divisor are first arranged according
DIVISION. 47
to descending powers of some common letter, a suppose;
and the quotient also is considered to be so arranged.
Then (Art. 62) the first term of the dividend will be the
product of the first term of the divisor and the first term
of the quotient; and therefore the first term of the
quotient will be jfimnd by dividing the fi^rat term of the
dividend by the first term of the divisor. If we now
multiply the whole divisor by the first term of the
quotient so obtained, and subtract the product from the
dividend, the remainder must be the product of the
divisor by the sum of all the other terms of the quotient;
and, this remainder being also arranged according to
descending powers of a, the second term of the quotient
will be fourid as before by dividing the first term of the
remainder by the first term of the divisor. If we now
multiply the whole divisor by the second term of the
quotient and subtract the quotient from the remainder, it
is clear that the third and other terms of the quotient can
be found in succession in a similar manner.
For example, to divide 8a" + Sa^b — 4!ab^ + 6' by 2a + b.
The arrangement is the same as in Arithmetic.
2a + 6 ) 8a' + 8a»6 + iiab^ + V ( W + 2ab + V
8a° + 4a»6
4a'6 + 4a6* + b^
4a*6 + 2a&*
2aV + 6"
2a6« + y
The first term of the quotient is 8a" 4 2a = 4a*.
Multiply the divisor by 4a' and subtract the product from
the dividend : we then have the remainder 4a 6 + 4a6'' + 6".
The second term of the quotient is 4a*6 ^ 2a = 2a6.
Multiply the divisor by 2a6, and subtract the product
from the remainder: we thus get the second remainder
2a6* + 6*. The third term of the quotient is 2a6* * 2a = 6*.
Multiply the divisor by 6*, and subtract the product from
48 DIVISION.
2a6'f 6', and there is no remainder. Since there is no
remainder after the last subtraction, the dividend must be
equal to the sum of the different quantities which have
been subtracted from it; but we have subtracted in suc
cession the divisor multiplied by 4a*, by +2a6, and by
+ 6* ; we have therefore subtracted altogether the divisor
multiplied by 4a' + 2db + b\ And, since the divisor mul
tiplied by 4a* 4 2ab + 6* is equal to the dividend, the
required quotient is 4a* + 2ab + 6*.
The dividend and divisor may be arranged according
to ascending instead of according to descending powers of
the common letter, as in the last example considered with
reference to the letter 6 ; but the dividend and the divisor
must both be arranged in the same way.
71. The following are additional examples:
Ex. 1. Divide a*a»6+2aVa6»+6* by a»+6».
a'^b + a^b^ a6» + 6*
 a'6  db^
+ a26« +6*
+ aaj2 4.54
Ex. 2. Divide a*+a«63+6* by a2a6 + 6».
a^ab+bAa* +a^b* \b^(a^+ab+b^
' a^a^b + a^b^ ^
+ a»6 +6*
•¥a^ba*b^+db^
In this example the terms of the dividend were placed apart, in
order that *like' terms might be placed under one another without
altering the order of the terms in descending powers of a. The
subtractions can be easily performed without placing *like' terms
under one another; but the arrangement of the terms according to
descending (or ascending) powers of the chosen letter ahoold never
be departed from.
Ex. 3. Divide o'+6'+c»3a6c by a+b+c.
DIVISION. 49
3o6c+6*+c*
ahc
m
2a6c+6'+c»
a6c oc*
—
Where, as in the ahove example, more than two letters are
involved, it is not sufficient to arrange the terms according to
descending powers of a; but h also is given the precedence over c.
By using brackets, the above process may be shortened. Thus
a+ 6+c ) a»  3a6c+ t'+c* ( o* a (6 +c) + (6*  6c+c')
^ a» + a2(&+c)__J
a»(&+c)~a(6+c)*
72. The method of detached coeflScients may often be
employed in Division with great advantage. For example,
to divide
2a?*7a;"+6a;* + 3aj'3a;' + 4a?4 by 2i»'3^' + i»2,
we write —
23 + 12)27 + 5 + 33 + 44(121 + 2
23+12
4+4+63+4
4+62+4
4
2 + 7
2 + 3
7 + 4
1 + 2
4
4
4
6 + 2
6 + 2
4
4
The first term of the quotient is a? and the other
powers follow in order : thus the quotient is
S. A. 4
60 DIVISION.
73. Extended definition of Division. In the
process of division as described in Art. 70, it is clear that
the remainder after the first subtraction must be of lower
degree in a than the dividend; and also that every re
mainder must be of lower degree than the preceding
remainder. Hence by proceeding far enough we must
come to a stage where there is no remainder, or else
where there is a remainder such that the highest power
of a in it is less than the highest power of a in the divisor,
and in this latter case the division cannot be exactly per
formed.
It is convenient to extend the definition of division to
the following : To divide A by B is to find an algebraical
expression Usuch that BxC is either equal to A, or differs
from A by an expression which is of lower degree, in some
particular letter, than the divisor B,
For example, if we divide a' + 3a6 + 46' by a + b, we
have
a + 6 ) a* + 3a6 + 46' ( a + 26
a'h ab
2a6 + 46'
2a6 + 26'
+ 26'
Thus (a' + 3a6 + 46') 4 (a + 6) = a + 26, with remainder
26' ; that is a' + 3a6 + 46' = (a + 6) (a + 26) + 26'. We have
also, by arranging the dividend and divisor differently,
6 + a ) 46' + 3a6 + a' ( 46  a
46' + 4a6
— a6f a'
— a6 — a'
+ 2a'
Hence a change in the order of the dividend and
divisor leads to a result of a different form. This is, how
ever, what might be expected considering that in the first
DIVISION. 61
case we find what the divisor must be multiplied by in
order to agree with the dividend so far as certain terms
which contain a are concerned, and in the second we find
what the divisor must be multiplied by in order to agree
with the dividend so far as certain terms which contain b
are concerned.
When therefore we have to divide one expression by
another, both expressions being arranged in the same
way, it must be understood that this arrangement is to
be adhered to.
74. Def. A relation of equality which is true for all
valves of the letters it contains, is called an identity.
The following identities can easily be verified, and
should be remembered :
(a? + 2cuc + a') r (a? + a) =saj + a.
{a^ —,2a!X + a'*) 5 (a? — a) = a? — a.
(a?* — a') r (a? ± a) = a? T a.
(a:' T a') 5 (a? ? a) = a? ± cw? + a*.
(a?* 'a*)'7{xT(i)=^ ± flw?' + a*a? + a*,
(a?* + a V + a*) ^ (a^ T aa? + a*) = aj' ± oa? + a*.
1.
Divide
2.
Divide
3.
Divide
4.
Divide
6.
Divide
a.
Divide
EXAMPLES III.
a^ 9y* by aj + 3y.
aj*16y* by aj»~4y».
27a» + 6 V by 4y + 3x.
Ba^ixy^y' by 2yx.
l5aJ*+4a^ by la?.
05*  5icy* + 4^* by ajj^
4—2
52 EXAMPLES.
7. Divide l6a^ + 5aj« by l2a; + a^.
8. Divide m*  6mn" + S/i* by m' — 2mn + n'.
9. Divide 1  7aj« + 6a^ by (1  x)\
10. Divide la? by laj«.
11. Divide l+aj8aj*+19a^15aj* by l + 3aj5ar».
12. Divide 4 9fc"+ 12aj» 4aj* by 2 + 3aj 2iB".
13. Divide 4aj*  9aj*2^" + 6ay  y* by 2aj« + 3a^y*.
14. Divide fic'3a:' + 3aj + y" 1 by aj + y1.
15. Divide os^ + 0?*^ + a^y' + ^^ + a^* + y" by oj* + a?y + y".
16. Divide
a8*5a5*y + 7a^y*«'^4a5y* + 2y' by «*  3a5'y + 3«y"  y*.
17. Divide a' 26"6c' + a6ac + 7ftc by a6 + 2c.
18. Divide a* + 26'  3c' + 6c + 2ac + 3a5 by a + 6c.
19. Divide
Qal" + 46*  <^6 + 1 3a6» + 2a'6' by 2a' + 46'  3a6.
20. Divide re* + y*  »* + 2iB'y' + 2«'  1 by a:* + 2/'«»+ I.
21. Divide a'3a'6+3a6'6'c» by a6c.
22. Divide a» + 86"  c" + 6a6c by a + 26c.
23. Divide
a* + 86» + 27c»18a6c by a' + 46' + 9c'  66c  3ca  2a6.
24. Divide 27a»  86'  27c»  54a6c by 3a  26  3c.
25. Divide ocoJ* + (oc?  6c) ic'  (oc + 6c^) a? + 6c by aa;6.
26. Divide
2a'fc' . 2 (6  c) (36  4c) y' + a6a:y by aaj+2(6c)y.
27. Divide
9a'6'  1 2a*6 + 36» + 2a'6' + 4aP  1 la6* by 36* + 4a'  2a6'.
28. Divide aj' + y* by as + y; and from the result write down
the quotient of (a? + y)' + »* by a? + y + «.
29. Divide aj* — y" by x — yy and hence write down the
quotient of (x + y)'  8»' by a? + y  2«.
CHAPTER VL
Factors.
75. DeflnitionB. An algebraical expression which
does not contain any letter in the denominator of any
term is said to be an integral expression : thus \a^b — ^6'
is an integral expression.
An expression is said to be integral with respect to any
particular letter^ when that letter does not occur in the
denominator of any term : thus — I — ^ is integral wiHi
•^ a a +
respect to x.
An expression is said to be rational when none of its
terms contain square or other roots.
76. In the present chapter we shall shew how factors
of algebraical expressions can be found in certain simple
cases.
We shall only consider rational and integral expres
sions ; and by the factors of an expression will be meant
the rational and integral expressions, or the expressions
which are rational and integral in some particular letter,
which exactly divide it.
77. Monomial Factors. When some letter is
common to all the terms of an expression, each teim, and
therefore the whole expression, is divisible by that letter.
Thus 2ax + a^=:x(2a+x),
ax + a*x^=s ox (1 + ax),
and 2a'6»a; + 3a«6»y = a'i* {2ax + Sby),
Such monomial factors, if there be any, are obvious on
inspection.
54 FACTORS.
78. Facton found by comparing with known
identities. Sometimes an algebraical expression is of
the same form as some known result of multiplication:
in this case factors can be written down at once.
Thus, from the known identity
we have
o»463=a3(26)2=(a+26)(a26),
a^2=a^(^2f={a+^2)(a^2),
a* 166<=(a«)« (463)2=(a3 + 462)(aa462)
=(a3 + 462)(a+26)(a26),
and a»9a6a=a(aa96*)=a(a + 36)(a36).
Again, from the identity
we have
a«+86»=a8+(26)3=(a+26){a3a(25) + (26)»}
= (a+26)(a82a6 + 46«),
8a» + 276« = (2a)' + (363)8 = (2a + Sb^ { (2af  (2a) (36») + (363)'}
= (2a + 36«) (4a«  6a6« + 96<) ,
and a»+a;»=(a8)3+ («»)»= (a»+«8)(a«a»a:8+a;«)
= {a+x){a^ax + x*){cfi'a'a^ + afi).
And, from the identify
a»"6«=(a6)(a3+a6 + 6«),
we have
a«6»  g«»y'= (ab  i^cy ) (a^b* + ^ a6xy + j a:v) •
The following are additional examples of the same
principle :
(i) (a+6)»((! + d)2={(a+6) + (c + cO} {(« + &) (c+d)}
= (a + 6 + c + d)(a+6cd).
(ii) 4a>6a(a>+68c2)2={2a6 + (aa+6»c3)}{2a6(aa+6ac3)};
and, since
2db+a^{'b^^c^={a+b)*<^={a+b + c){a+bc),
and 2a6o«6Hc3c»„(a_5)j_(c+a_j)(ca+6),
we have finally
4a26«(a2 + 62c«)«=(a+6 + c)(6 + ca)(c + a6)(a + 6c).
FACTORS. 65
(iii) (a+26)8(2a46)«
= {(a + 26)(2a + 6)}{(a+26)2+(a+26)(2a + 6) + (2a+6)3}
=:(6o)(7aa+13a6 + 762).
79. Facton of a^+pos + q found by inspection.
From the identity
{x + a) {x + b) =^ a^ + {a i b) X + ab,
it follows conversely that expressions of the form
a^ {px + q
can sometimes, if not always, be expressed as the product
of two factors of the form x + a, x + b.
We shall presently give a method by which two factors
of x^ipx + q of the form x + a and x + b can always be
found ; but whenever a and b are rational, the factors can
be more easily found by inspection. For, if (x + a) {x + b),
that is a^+ (ahb) x{ ab, is the same as x^ +px + q, we
must have a + b—p and ah = q. Hence a and b are such
that their sum is p, and their product is q.
For example, to find the factors of sc^h 7x + 12. The factors will
be a; + a and a: + 6, where a + 6 = 7 and ab = 12. Hence we must find
two numbers whose product is 12 and whose sum is 7: pairs of
numbers whose product is 12 are 12 and 1, 6 and 2, and 4 and 3;
and the sum of the last pair is 7. Hence a^ + 7a; + 12 = (a;+ 4) (x + 3).
Again, to find the factors of a;'  7a: + 10. We have to find two
numbers whose product is 10, and whose sum is 7. Since the
product is +10, the two numbers are both positive or both negative;
and since the sum is  7, they must both be negative. The pairs of
negative numbers whose product is 10 are ~ 10 and  1, and  5
and  2; and the sum of the last pair is  7. Hence a:^  7a;+ 10=
(aj6)(a;2).
Again, to find the factors of a:^+3a;18. We have to find two
numbers whose product is  18 and whose sum is 3. The pairs of
numbers whose product is  18 are  18 and 1,9 and 2,6 and 8,
.  8 and 6,2 and 9 and  1 and 18 ; and the sum of 6 and  8 is 8.
Hence a? + 8aj  18 = (a: + 6) (a;  8).
It should be noticed that if the factors of x^+px+q bo
xha and a? + 6, the factors of a^ +pxy + qy* will be a? + ay
and x\by\ also the factors of {x + yY + p {x ■\ y) z •{ qz^
will hQ x + y^az and x + y + bz.
66 FACTORS.
Hence from the above we have
a;^+7icy + 12y«s=(a;+4^)(a;+3y),
a;> + Sxy^  18y*= {x + 6y«) {x  3y2),
(a+6)37(a + 6)a:+10«*=(a + 66a;)(a + 62a;),
and «*6a;« + 4 = («»)a6a;a+4 = (a;»4)(a:al)
= (a;+2)(a;2)(a; + l)(a;l).
EXAMPLES IV.
Find the factors of the following expressions :
1. a* 166*. 2. 16a;*81aV.
3. 16(3a^26)«. 4. 4/  (22J  a;)».
5. 20aV45aa^. 6. 36aV4aVy*.
7. (3a»  6')'  (a«  36»)». 8. (5a»  36«)«  (3a«  bhy.
9. (5a:» + 2a;  3)»  (aj»  2a;  3)».
10. (3a;«  4a;  2)»  (3a;» + 4a;  2)«.
11. 32aV46^ 12. (a«  26c)'  86V.
13. a«2a8. 14. a;+12a;».
15. 1 18a; 63a;". 16. 8a4a»4.
17. a»6  4a V + 3a6». 18. a*6 + 5a»6' + 4a V.
19. (6 + c)»  6a (6 + c) + ba\
20. 9(a + 6)»6(a + 6)(c + (/) + (c + rf)«.
21. a;*29a;»+100. 22. 100a;*  29a;'y* + y*.
23. a;*8a;»/»*+162/V. 24. 9a' 10aV + a«6\
25. a;"2aa;6* + 2a6. 26. a;" + 2a;2/  »*  2ay.
27. 4(a6 + crf)»(a» + 6»c«rf7.
28. 4 (a;y  a6)' (a;» + y«  a«  67.
FACTORS. 57
80. Facton of general quadratic ezpreMion.
We proceed to shew how to find the factors of any ex
pression of the second degree in a particular letter, a
suppose.
The most general quadratic expression [Art. 60] in w
is cwj" + 6a? + c, where a, b and c do not contain x.
The problem before us is to find two factors which are
rational and integral with respect to x, and are therefore
each of the first degree in x, but which are not necessarily^
and not generally, rational and integral with respect to
arithmetical numbers or to any other letters which may
be involved in the expression.
The method of finding the factors of oof + 6a? + c con
sists in changing it into an equivalent expression which
is ihs difference of two squares.
We first note that since a? + 2aa? + a' is a perfect
square, in order to complete the trinomial square of which
of and 2ax are the first two terms, we must add the
square of a, that is, we must add the square of half the
coefficient of x,
^Qit example, x^\hx\& made a perfect square, namely ( ^ + n ) »
by the addition of ( ^ j ; also t^^x is made a perfect square, namely
Ix  ly, by the addition o^ ( f )' = j •
81. To find the factors of ao? ■\^hx\c.
aafhbx + c = a(a^+ a?H   .
\ a aj
Now aj*+  a? is made a perfect square, namely (x + ^j y
by the addition of f ^ j = ji • And, by adding and sub
b*
tracting / to the expression within brackets, we have
68 FACTOBS.
= a •](« +
Hence as the difference of any two squares is equal to
the product of their sum and difference, we have
cu^ ^bx + e
''r + 2a + V 4a' J r^2a V W ]'
Thus the required factors have been found*.
Ex. 1. To find the factors of ac* + 4a; + 8.
a;»+4a; + 3=a;3+4aJ + 44 + 3 = (a; + 2)al = (a; + 2 + l)(a; + 2l)
= (ae + 3)(a5 + l).
Ex. 2. To find the factors of x^  5x + 3.
^5..a=^5..(.)'()%8=(.)'LB
=(*!+ \/t)(*§\/t)
Ex. 3. To find the factors otZa^^ + l,
M('i)*i)Hi*l)(M)(i)'')
Ex. 4. To find the factors of x" + 2aa;  6»  2ah.
aj»+2aaj6«2a6=a:« + 2aa; + a»a»6»2a6=(a;+a)«(a46)'
= {a: + a + (a+6)}{a5+a(a + 6)} = (aj + 2a+6)(a;6).
* It will be proved later on [see Art. 91] that an expression containing
a can be resolved into only one set of factors of the first degree in x.
FACTOBS. 59
82. Instead of working out every example from the
beginning we may use the formula
oar* + bx + c
and we should then only have to substitute for a, b and o
their values in the particular case under consideration.
Thas to find the factors of 3a:'4a;+l. Here a=3, 6=4, c=l.
„ ^ /b^iac 716  12 /I 1 ., . .
^^"^ \/Ta^'^ V "36~= V 9 = 3' ^^'^ e»P«8«>on is
therefore equivalent to 8 (*o + q){^o " 5) = ^(*~i) (*~^*
83. We have from Art 81
aa^^bx + c
Now, for particular values of a, b, c, — 75 — may be
positive, zeroy or negative.
I. Let — jg — be positive. Then the two factors of
cut? + bx + o will be rational or irrational according as
— 7T — IS or IS not a perfect square.
II. Let . . , — be zero. Then
Hence aa? + bx + c is a, perfect square in x, if 6' — 4ac = 0.
"^ ^ac
III. Let — yj — be negative. Then no positive or
negative quantity can be found whose square will be equal
to — 78— ; for all squares, whether of positive or nega
tive quantities, are positive.
60 FACTORS.
Expressions of the form V— a, where a is positive, are
called imaginary, and positive or negative quantities are
distinguished from them by being called reai.
We shaU consider imaginary quantities at length in a
subsequent chapter : for our present purpose it is sufficient
to observe that they obey all the fundamental laws of
Algebra; and this being the case, the formula of Art. 81
will hold good when 6' — 4ac is negative.
Note. For some purposes for which the factors of
expressions are required, the only useful factors are those
which are altogether rational: on this account irrational
and imaginary factors are often not shewn. Thus, for
example, the factorisation of ^ — 8 is for many purposes
complete in the form (a? — 2) («* + 2a? + 4) *, the imaginary
factors of «* + 2a? + 4, namely
a? + l + V^ and a?+lV^,
not being shewn.
84. We have in Art. 81 shewn how to resolve any
expression of the second degree in a particular letter into
two factors (real or imaginary) of the first degree in that
letter.
It should be noted that the factors of the most general
expression of the third degree, or of the fourth degree,
can be found, although the methods are beyond the range
of this book ; expressions of higher degree than the fourth
cannot however, except in a few special cases, be resolved
into factors.
85. Facton fbund by rearrangement and
grouping of terms. The factors of many expressions
can be found by a suitable rearrangement and grouping
of the terms.
For example
l+<MJa;3aaH»=l + aaja?»(l + (M;) = (l+(M;)(la;2)
= (l + ox)(l+a?)(la:);
♦ The reason of this wiU appear from Art. 178 and Art. 192.
FACTORS. 61
or we may write the expression in the form
and the £aotors 1  oc*, 1 + ax are now obvious.
For the best arrangement or grouping no general rule
can be given : the following cases are however of frequent
occurrence and of great importance.
I. When one of the letters occurs only in the first
power, the factors often become obvious when the expres
sion is arranged according to powers of that letter.
Ex. 1. To find the factors otab^he+ed+da.
Arranged according to powers of a we have a(b+d) + be+ed,
which is at once seen to hea{b{d^+e(h\d)={a{c){b+d),
Ex. 2. To find the factors of x'+(a+&+c)st;+a5 + ac.
The expression s a (s + 6 + c) + as* + &« + ex = (a + as) (s + 5 + c).
Ex. 8. To find the factors of ox* +as+ a+ 1.
aa5»4aB+a+ l=a(a5»+ 1) +«+ 1= («+ 1) {a(iB*  «+ 1) + 1}.
Ex. 4. To find the factors of a*+2ah'2ac9b^+2bc.
The given expression is of the first degree in c; we therefore write
it in the form a'+2a&>3&>2c(a5)
»(a6)(a+8&)26(a5) = (a&)(a+8&2c).
II. When the expression is of the second degree with
respect to any one of the letters; factors, which are rational
and integral in tiiat letter, can be found as in Art. 81.
Ex. 1. Find the factors of a* + Sb*  c* + ^be  4tab.
Arranging according to powers of a, we have
a«4a6+86>c*+26c=a»4a6+462.46a+862c» + 26c
= (a26)«(6c)»={(a26) + (6c)}{(a26){6c)}
= (a5(j)(a86+c).
Ex. 2. Find the factors of a«6»c«4d*2(ad6c).
The expression
=o>2ad6«c«+«P+26c
=:o«2ad+ci»6»c>+26c=(od)»(6c)»
=s(a~d+6c)(od6+c).
62 FACTOBS.
Ex. 3. Findthefaotoraof a>+2a&ac3&3+5dc2(^.
The expression
=a«+a(26c)36a+56c2c«
=a>+a(26c)+ ^^^Il^V  (^^V  36» + 66c  2c2
= ^a + ?5^y  i {462  46c + c» + 126«  206c + 8c3}
= f+?^%(«8c)}f.?^(«8o)}
= (a+362c)(a6+c).
Ex. 4. Find the factors of «* + «'  2ax + 1  a'.
Arranging according to powers of a, we have
{o«+2aa;laj*a?*}={aa+2ax+a;2l2a;»a:*}
=  {(a + a;)> (1 +a:')»} =  (a+aj+ 1 + a*) (a+«  1  a;*).
III. When the expression contains only two powers
of a particular letter and one of those powers is the square
of the other, the method of Art. 81 is applicable.
Ex. 1. To find the factors of ar^lO^c^+O.
«*ia«»+9=«*10«»+2626+9=(a:"6)«16
= (a;««6+4)(«»64)=(«>9)(a?"l) = (fl5+3)(aj3)(a?+l)(ajl),
or thus:— «*10aj»+9=(aJ*+3)«16aj»
=(a?+3+4a;)(»«+84a;)=(a; + 3)(aJ+l)(«3)(a;l).
Ex. 2. To find the factors of a^+a^+ 1.
Two real quadratic factors can be found as foUows:
««+a!>+l=(a;9+l)«a;9=(a^+l + a?)(iB3+la;).
Ex. 8. To find the Caotors otafi 28aB>+ 27.
a«28«»+27=««28a»+14«14a+27=(a!«14)a13>
= («»■ l)(a>27)=(aJl)(«8)(a!«+«+l)(x«+8a;+9).
In this ease, and also in Ex. 1, two factors can be seen bv
inspection, as in Art. 79*
FACTORS. 63
Ex. 4. To find the factors of a*+6*+c*262c«2<jaa*2a«6«.
Arranging according to powers of a, we have
a*  2a«(6a+ca) + 6* + c* 26»c«
= {a»(6«+c*)}«46«c>=(a>6>c826c)(a«62_c«+26<j)
= {a«(6 + c)«}{a>(6c)2}
= (a+6+c)(a6c)(a6+c)(a + 6c).
IV. Two factors of aP* ^bP + c, where P is any
expression which contains ^, can always be found by the
method of Art. 81 ; for we have
Ex. 1. To find the factors of (aj»4a;)'+4(a;«+a;)  12.
Since P«+4P12=(P2)(P+6),
the given expression = (as? + »  2) («* + a; + 6)
= (a;+2)(«l)(a:"+aj+6),
the factors of a^+«+6 being imaginaiy [see Art 83, Note].
Ex. 2. To find the factors of (a:?+«+4)*+8x(a?»+a;+4) + 16«3.
The given expression = {(x' + aj + 4) + 8a;} {(oc* + at + 4) + 6«}
= («» + 4a5 + 4)(ar» + 6a! + 4)
= (aj+2)«(aj2 + 6»+4).
Ex. 8. To find the factors of
2(a:3+6aj+l)«+6(«»+6a; + l)(a;« + l) + 2(«»+l)«.
Since 2P« + 6PQ + 2Q»= (P + 2Q) (2P + Q),
the given expression
= {(«»+6«+l) + 2(a^ + l)}{2(ar» + 6«+l)+»«+l}
= (So? + 6a5 + 8) (8a!» + 12a! + 8)
=9(aj+l)>(a5«+4aj+l).
Ex. 4. To find the factors of (aJ"+aj+l)(aB?+a5+2)12.
The given expression = («^ + x)' + 8 (ac^ + a;)  10
= (ar»+x2)(«a+a; + 6)
= (x + 2)(xl)(x«+x + 6).
^
iS
62 FACTOBS.
Ex. 3. Findthefaotoraof a* +2a& 00863+52)020^.
The expression
=a«+a(2&c)36« + 56c2o«
 ^a + ?^ y  J {462  46c + c' + 126«  206c + Sc^}
/ 26 cV 1/.^ o X5
f 26c 1,,^ „ ,> I 260 1... „ J
= (a+362c)(o6+c).
Ex. 4. Find the factors of «* + «'  2aaj + 1  a*.
Arranging according to powers of a, we have
{a«+2aa;laj*a^}={aa+2ax+«*l2a;2a:*}
=  {(a + a;)« (1 + «•)•}= (a+a5+ 1 + aJ*) (a+« 1 a;*).
III. When the expression contains only two powers
of a particular letter and one of those powers is the square
of the other, the method of Art. 81 is applicable.
Ex. 1. To find the factors of a:*  lOa;^ ^ 9,
«*ia«»+9=«*10«»+2626 + 9=(a:"6)«16
= (a:t«6+4)(a»64)=(«>9)(a?l)=:(aj+8)(aj8)(a?+l)(ajl),
or thus:— «*10aj»+9=(aJ*+8)«16a:"
= {a?+S+4x){x*+S'4x)={x + S){x+l){xS)(xl).
Ex. 2. To find the factors of a:^+:B*+l.
Two real quadratic factors can be found as foUows:
x^+sfi+l={x^ + l)^ay^={aP+l + x){afl^lx).
Ex. 8. To find the factors of s«  28aB> + 27.
a«28«» + 27=««28a»+14«14a + 27=(a!«14)«lS>
= (a»l)(a>27) = (»l)(«8)(a»+a; + l)(a;« + 8a:+9).
In this ease, and also in Ex. 1, two factors can be seen bv
inspection, as in Art. 79«
FACTORS. 63
Ex. 4. To find the factors of a*+6*+c*262c«2<jao>2a«6«.
Arranging aooording to powers of a, we have
a*  2a«(62+c2) + 6*+c*  26»ca
= {a*(ft*+c*)}»46»c»=(o«6«c«26c)(a«.6«c«+26<j)
= {a«(6 + c)a}{aa(6c)»}
= (a + 6 + c)(a6c)(a6 + c)(a + 6c).
IV. Two factors of aP^ ^bP + c, where P is any
expression which contains a?, can always be found by the
method of Art. 81 ; for we have
aF' + bP + c
Ex. 1. To find the factors of (a^\x)^+^{x*+x) > 12.
Since P«+4P12=(P2)(P+6),
the given expression = (as? + a:  2) (a;* + a? + 6)
zz{x+2){xl)(x*+x+6),
the factors of a^+x+6 being imaginary [see Art 83, Note].
Ex. 2. To find the factors of (aB^+a?+4)*+8x(a?»+a?+4) + 16a^.
The given expression = {(x"+aj+ 4) + 3«} {(it*+ at + 4) + 6«}
= (aj» + 4aj + 4)(aj" + 6a5 + 4)
= (aj+2)«(a:a+6a?+4).
Ex. 8. To find the factors of
2(x«+6aj+l)«+6(ic» + 6a; + l)(a;« + l) + 2(a^+l)a.
Since 2P« + 6PQ + 2<2» = (P + 2Q) (2P + Q),
the given expression
= {(a;» + 6«+l) + 2(aj3 + l)}{2(ar» + 6«+l)+ai» + l}
= (3aj?+ 6a5+8) (3a?+ 12aj+ 8)
= 9(a;+l)«(a5«+4aj+l).
Ex. 4. To find the factors of (aJ*+«+l)(ai^+as+2)12.
The given expression = {x^ + x)* + 3 (ac^ + as)  10
= (a:a+a!2)(«a + » + 6)
= (a? + 2)(a?l)(a;»+a: + 6).
7^
64 EXAMPLEa
EXAMPLES V.
Find the factors of the following expressions :
1. Q? + aa? — 05 — a.
2. ac " hd — ad ■\ he,
3. ac^ + h<Pad'M.
4. acoi? + (pc + ad) xy + hdif,
5. oca? + hco^ + adx + 6c?.
6. (a + 6)« + (a + c)«(c + (/)»(6 + rf)«.
7. a^j^a%'fih''h\
8. a*a"6a6» + 6*.
9. aVa»6« + l.
10. a^y'"a^»»y*»« + «\
11. a;V»*  a5*»  2/*^ + 1.
12. aj^ + rc'y + aj^^ + y*".
13. aj(aJ + «)y(y + »).
14. a;*~7«'18.
15. aj*  23a:' + 1.
16. aj*14a^y' + y*.
17. »• + »*+!.
18. a;*2(a'46*)a:« + (a'57.
19. «*4a^/««+4yV.
20. aJ»2(a + 6)aja6(a2)(6 + 2).
21. «* + &B" + aaj + a6.
22. (l+3^)»2a^(l+2/') + aJ*(ly)*.
23. as*  y*  3«"  2a» + 4y«.
FACTORS. 66
25. a»  36'  3c» + 106c  2ca  2ah.
26. 2a'  7a6  226«  5a + 356  3.
27. l + (6a»)a"a6aj».
2a l2aaj(ca")a" + ac«".
29. a'(6  c) + 6«(c  a) + c"(a  6).
30. 6'c + 6c* + c'a + ca* + a*6 + a6* + 2a6c.
31. a'6  a6* + a*e  oc" 2a6c + b'c + hc\
32. af{a+l)xy(X'y) {ab)'^^{b + 1).
33. aaj(2^ + 6») + 6y(6iB« + aV).
84, 2«" ~ iic'y  «■« + 2ajy" + 2a5y»  y *«.
35. xyz (a* + y* + «")  y*«*  sfa?  ic'y'.
36. (iB"+ a:)"  14 (a:» + a:) + 24.
37. (i»» + 4aj+ 8)» + 3aj(iB* + 4aj + 8) + 2a^.
38. (aj+l)(aj + 2)(aj+3)(aj + 4)24.
39. (x+l) (a; + 3) (aj+ 5) (a: + 7) + 15.
40. 4(aj+5)(aj + 6)(aj+10)(a:+12)3a;». ^
86. Theorem. The eaypression aS* — a" is divisible by
os^a, for all positive integral valvss of n.
It is known that a? — a, x^ — a^ and a?" — a' are all
divisible by a? — a.
We have «?* — a* = a?" — oa?"'* + oa?*"^ — a*
= ar"'(a?  a) + a (a?""'  a*"').
Now if OB — a divides a^~* — a""^ it will also divide
aj*"* (a? — a) + a (a?"'* — «*"*)> ^^^^^ ^'*> i* will divide a?* — a*.
Hence, if x — a divides a;*'* — a"~* it will also divide
of* a''.
S.A. 5
66 FACTORS.
But we know that x — a divides x* — a*; it will therefore
also divide x* — a*. And, since x — a divides x* — a* it will
also divide a^ — a^ And so on indefinitely.
Hence a;* — a** is divisible by x — a, when n is any
positive integer.
87. Since a;~ + a* = i»~  a~ + 2a* it follows from the
last Article that when x"^ + a** is divided by a? — a the
remainder is 2a'*, so that x* + a" is never divisible by ^ — a.
If we change a into — a, x^a becomes a? — (— a) = a? + a;
also a?" — a" becomes a?" — (— a)**, and a;* — (— a)** is a?" + a**
or a?" — a" according as n is odd or even.
Hence, when n is odd
x"" + a** is divisible hy x\a,
and when n is even
a?" — a* is divisible by ^ + a.
Thus, n being any positive integer,
X — a divides x"  a* always,
a; — a „ a?" + a" never,
a? + a „ a?* — a" when n is even,
and x + a „ x*" + a* when n is odd.
The above results may be written so as to shew the
quotients: thus
«» ^«
•? ?_ _ ^»l __ «,^« ^ I . «,»8 ^8 , , ^«I
x — a
= a;"*' + a;""* a + a?""* a' + 4a*
f^J? = aj*^~aj^»a + a?"'a» ±a"*S
a? + a
the upper or lower signs being taken on each side of the
second formula according as n is odd or even.
88. Theorem. If any rational and integral eospres
sion which contains x vanish when a is put for x, then will
x — abe a factor of tlie expression.
FACTORS. 67
Let the expression, arranged according to powers of x,
be
cw?* + 6a;*"* + ca?*"" +
Then, by supposition,
aa* + 6a*"* + ca*"" + =0.
Hence ax*\ 6a?""* + ca;*"' +
= aa;* + 6a?*"* + ca?*"* +  (aa*+6a*"* + ca*"'+ ...)
= a (a?*  a*) + 6 (a?*"*  a""*) + c (a?*"*  a*"') +
But, by the last Article, a?*  a*, a?*'*  a*"*, a?*~*  a*"*,
&c. aie all divisible by a? — ct
Hence also oa?* + 6a?*"* + ca?*"* + is divisible by
a? —a.
The proposition may also be proved in the following
manner.
Divide the expression aa?*+ 6a?*"* + cx'''^ + by a? — a,
continuing the process until the remainder, if there be any
remainder, does not contain x ; and let Q be the quotient
and jB the remainder.
Then, by the nature of division,
oa?* + 6a?*"' + ca?*"" + =Q(^a) + B,
and this relation is true for all values of a?.
Now since R does not contain a?, no change will be
made in JB by changing the value of x : put then a? = a, and
we have
aa» + 6a** + ca*»+ = Q («  a)^R = R
Hence, if any expression, rational and integral in x
be divided by x — a, the remainder is equal to the result
obtained by putting a in the place of x in the expression.
It therefore follows that the necessary and sufficient
condition that an expression rational and integral in a?
may be exactly divisible by a? — a is that the expression
should vanish when a is substituted for a?.
5—2
68 FACTORS.
Ex. 1. Find the remainder when aj^ _ 4^2 + 2 is divided by a:  2.
The remainder =2»4.22 + 2= 6.
Ex. 2. Find the remainder when 3?  2a^x + a' is divided by a?  a.
The remainder is a'  2a* + a'=0, so that afi  2a*a; + a' is divisible
by 05  a.
Ex. 8. Shew by substitution that x1, a; 5, a;+2 and a; +4 are
factors of aj*  23a:>  18a; + 40.
Ex. 4. Shew by substitution that a  & is a factor of
Put a=b and the expression becomes a*{ac)+a*{ca), which
is clearly zero: this proves that a & is a factor.
Ex. 5. Shew that a is a factor of
(a+6+c)«(a+6+c)'(a6+c)'(a+6c)'.
89. We have proved that a? — a is a factor of the
expression cm?" + 6a?*** + cw?*"* + , provided that the
expression vanishes when a is put for a?.
If the division were actually performed it is clear that
the first term of the quotient, which is the term of the
highest degree in x, would be cuc^'K Hence the given ex
pression is equivalent to
(a? — a) (cM?*'* + &c ).
Now suppose that the given expression also vanishes
when x=fi\ then the product of a? — a and 00;*"* +
will vanish when ic = l3; and since x—a does not vanish
when x = l3, it follows that cw?""*+ must vanish
when a?=/8. Hence a? — ^8 is a factor of aa?*"* + &c.; and,
if the division were performed, it is clear that the first
term of the quotient would be oaj""".
Hence the original expression is equivalent to
(xa) {x  ^) (aa?*"» + &c ).
Similarly, if the original expression vanishes also for
the values 7, S, &c. of x, it must be equivalent to
(a? a) (a; /3) (a? 7) (a? S) (aa?»^ + &c ),
FACTORS. 69
where r is equal to the number of the factors a? — a,
a?  /8, &c.
If therefore the given expression vanishes for n values
a, ^, 7, &c. there wUl be n factors such as a; — a, and the
remaining factor, ax*'^ + &c. will reduce to a; and hence
the given expression is equivalent to
a (a? — a) (a? — /8) (pc — y)
Cor. If any of the factors a? — a, a? — /8, ... occur more
than once in ax*+bar'^+ ... , it can similarly be proved
that the expression is equivalent to a{x — OLf{x—Sf...,
the factors a? — a, x — ^, .,, occurring respectively p, q, ...
times, and p + q+ ... = n,
90. Theorem. An expression of the nth degree in x
cannot vanish for more than n values of x.
For if the expression
cw?" + 6a?*"^ + ca?""* +
vanishes for the n values a, A 7 , it must be equivalent
to
a{x " d) (x — &){x — y)
If now we substitute any value, h suppose, different
from each of the values a, /8, 7, &c.; then, since no one of
the factors A? — a. A; — ^, &c. is zero, their continued product
cannot be zero, and therefore the given expression cannot
vanish for the value x — hy except a itself is zero.
But, if a is zero, the original expression reduces to
6a?*"^ + cic""' + , and is of the {n — Vf degree; and
hence as before it can only vanish for n — 1 values of x,
except h is zero. And so on.
Thus an expression of the nth degree in x cannot
vanish for more than n values of a?, except the coefficients of
all the powers of x are zero; and when all these coefficients
are zero, the expression will clearly vanish for all values
of a?.
70 FACTORS.
91. Theorem. If two expressions of the nth degree
in X he equal to one another for more than n values of x,
they mill be eqvxil for all valves of x.
If the two expressions of the nth degree in x
flWJ** + fta?""^ + ca?"~' + ,
and ^d?** + q'a?*'^ + ra?"~"+ ,
be equal to one another for more than n values of x, it
follows that their diflference, namely the expression
(a p) a?" + (6  2) a?*"^ ^{pr) a?""' + ,
will vanish for more than n values of x.
Hence, by Art. 90, the coeflScients of all the different
powers of x must be zero.
Thus a— 1)=0, 6 — g = 0, c — r = 0, &c.
that is, a=p, l)=^qy c = ry &c.
Hence, if two expressions of the nth degree in x are
equal to one another for more than n values of a?, the
coefficient of any power of x in one expression is equal to the
coefficient of the same power of x in tiie other expression.
When any two expressions, which have a limited
number of terms, are equal to one another for all values
of the letters involved, the above condition is clearly
satisfied, for the number of values must be greater than the
index of the highest power of any contained letter.
Hence when any two expressions, which have a
limited number of terms, are equal to one another for all
values of the letters involved in them, we may equate the
coeffi^ents of the different powers of any letter.
92. Theorem. A rational integral expression con*
taining x can be resolved into only one set of factors of the
first degree in x.
For, if it be possible, let the expression ax"" + fta?""^ + ...
be equivalent to
a{x— OLf (x — ^y ,.., and also to a (x ~ ^f {x — y)"^ ...
FACTORS. 71
Put a? = a in both expressions ; then a (a — f )' (a — 17)"*. . .
must vanish, and therefore one at least of the quantities
^, f), ... must be equal to ct Let f = a ; remove one
factor x — a from both exjHressions, and proceed as before.
We thus prove that every factor of one expression occurs
to as high a power in the other expression; the two ex
pressions must therefore be identical.
93. Cyclical order. It is of importance for the
student to attend to the way in which expressions are
usually arranged. Consider, for example, the arrange
ment of the expression bc + ca + db. The term which does
not contain the letter a is put first, and the other terms
can be obtained in succession by a cyclical cha/nge of the
letters, that is by changing a into 6, b into c and c into a.
In the expression a* {b — c)\ J* {0 — 0) + a? {a — 6) the same
arrangement is observed; for by making a cyclical change
in the letters of a' (6 — c) we obtain 6' (c — a), and another
cyclical change will give c' (a — b). So also the second and
third factors of {b — c) {c — a) (a — b) are obtained from the
first by cyclical changes.
94. Symmetrical ezpressloiui. An expression which
is unaltered by interchanging any pair of the letters which
it contains is said to be a symmetrical expression. Thus
a + b + c, 6c + ca+a6, a' + 6* + c** — 3a6c are symmetrical
expressions.
Expressions which are unaltered by a cyclical change
of the letters involved in them are called cyclically sym
metrical expressions. For example, the expression
(6 — c){c — a) (a — b)
is a cyclically symmetrical expression since it is unaltered
by changing a into 6, b into c, and c into a.
It is clear that the product, or the quotient, of two
symmetrical expressions is symmetrical, for if neither of
two expressions is altered by an interchange of two letters
their product, or their quotient, cannot be altered by such
interchange.
72 FACTORS.
It is also clear that the product, or the quotient, of
two cyclically symmetrical expressions is cycUcally sym
metrical.
Ex. 1. Find the factors of a^(b '•c) + b^(ca) + c^ {a  6).
If we put hsscin the expression
a^^b'^c) + b^ca) + c*(ah) (i)
it is easy to see that the result is zero.
Hence bc is a factor of (i), and we can prove in a similar
manner that ea and a—b are factors.
Now (i) is an expression of the third degree; it can therefore only
have three factors.
Hence (i) is equal to
L{bc){c^a){ab) (u),
where L is some tmmberj which is always the same for all values of
a, bf c.
By comparing the coefficients [See Art. 91] of a' in (i) and (ii) we
see that X«=l.
We can also find L by giving particular values to a, & and c.
Thus, let a=0, 5 = 1, c=2; then (i) is equal to 2, and (ii) is equal
to 2L, and hence as before L=l,
Ex. 2. Find the factors of a^(be) + h^(ca) + c^(ab).
As in the preceding example, {bc), (ca) and {ab) are all
factors of
a^(bc)hb^{ca)+c^{a'b) (i).
Now the given expression is of the fourth degree ; hence, besides
the three factors already found, there must be one other factor of
the first degree, and this factor must be symmetrical in a, b, c, it
must therefore be a + & + c.
Hence the given expression must be equal to
L{b'c){ca){ab){a+b+c) (ii),
where Z is a number.
By comparing the co^cients of a^ in (i) and in (ii) we see that
L= 1; hence
a8(6c) + 68(ca)+c»(a6)=(6c)(ca)(a6)(a+6+c).
We can also find L by giving particular values to a, &, and e.
Thus, let a=0, &=1, c=2; then (i),i8 equal to 6 and (ii) is
equal to 6L, so that Z« =  1.
We may also proceed as follows:
Arrange the expression according to powers of a ; thus
a* (&  c)  a (63  c«) + Z/c (62  c2).
FACTORS. 73
It is now obvious that &  c is a factor, and we have
(hc) {a»a(62+6c+c«) + 6c(6 + c)}
= (6c)(ca){6*+6ca'ac}=(6c)(ca) (a6)(a + 6+c).
Ex. 3. Find the factors of li^c^ (bc) + cV (c  a) + a'ft^ (^ » 5).
By patting d=e in the expression
h^c^{he) + cH^(ca) + a%^{ab) (i),
it is easy to see that the result is zero ; hence &  c is a factor of (i).
So also ca and ab are factors.
The given expression being of the fifth degree, there mast be,
besides tiie three factors hc, ca, a~h, another factor of the
second degree ; also, since this factor must be symmetrical in a, 6, c,
it mnst be of the form L (a* + 6' + c^) + Jlf (6c + ca + ah).
Thus (1) is equal to
{hc) (ca) {ah) {La^+Lh^ + Lc'^ + Mhc^Mca^Mah} ...(ii).
Equating coefficients of a^ in (i] and in (ii) we see that L=0;
and then equating coefficients of 2rc* we see that M=  1. Hence
(i) is equal to
{hc) {ca) {ah){hc + ca+ah).
We may also proceed as follows.
Arranging according to powers of a, the factor hc which does
not contain a becomes obvious ; then, arranging according to powers
of h, the factor ca which does not contain h becomes obvious; and
so on. Thus
6V(6c)aa(6«c») + a*(62c2)
= (6c){6Va«(6« + 6<j + c«) + a»(6 + c)}
= {hc) {h*{c*  a^) + a%{ac)+a^ (ac)}
= {hc){ca){b^{c + a)a%a\}
:={hc){ca){{h*a^)c + h*aa^h}
ss  (6  c) (c  a) (a  h) {bc + ca + ah).
EXAMPLES VI.
Find the factors of the following expressions :
1. {yzy + {zxy + (xyy.
2. {yzy^(z^xy^{xyy.
3. a' (6»  c») + 6* (c«  a') + c' (a»  h').
4. a(bcy + h(cay + c{ahY.
74 EXAMPLES.
6. a(6c)' + 6(ca)» + c(a6)^
6. be (b—c) + ca(C'a) + ah (a — 6).
7. 6V(6c) + cV(ca) + aV(a6).
8. a* (6  c) + h" {ca) + c" (ab).
9. a' (hc) + b' (c  a) + c' (ab).
^ 10. (a + 6 + c)*(6 + ca)'(c + a~6)»(a + 6c)l
11. (a + 6 + c)**  (6 + c  a)'  (c + a  6)*  (a + 6  c)*.
12. a (6 + c  a)* + 6 (c + a  6)' + c (a + 6  c)"
+ (6 + c  a) (c + a — 6) (a + 6  c).
13. a' (6 + c  a) + 6' (c + a  6) + c* (a + 6  c)
 (6 + c — a) (c + a  6) (a + 6  c).
14. (b i c  a) (c + a  h) (a + b  c) + a {a — b + c) (a + b  c)
+ 6(a + 6c)(a + 6 + c) + c(a + 6 + c)(a6 + c).
15. (6  c) (a  6 + c) (a + 6 ~ c)+ (c  a) (a + 6  c) (a + 6 f c)
+ (a  6) ( a + 6 + c) (a  6 + c).
16. (x + yhzfa^y^ z\
17. (a; + y + »)*  a;**  y*  »^
18. (b c) (b h cY + {c a) {c + ay + (ab) (a + by,
19. (6c)(6 + c)* + (ca)(c + a)» + (a6)(a^6)^
20. (6c) {b + cy+{ca) (c + a)* + (a  6) (a + 6)*.
21. a*+6' + c^+5a6ca(a6)(ac)6(6c)(6a)
— c (c — a) (c — 6).
22. a'(a + 6)(a + c)(6c)+6'(6 + c)(6+a)(ca)
+ c' (c + a) (c + 6) (a  6).
23. (y + «)(» + c») (a? + y) + a;y2J.
24. a'(b+cy + 6' (c + a)» + c' (a + 6)» + a6c (a + 6 + c)
+ (a* + 6' + c') (6c + ca + a6).
25. (a: + y + 2;)*(y + 25)*(» + a:)*(aj + y)* + a;* + y* + «*.
26. a'(6 + c  2a) + b' (c + a 26) + c*(a + 6  2c)
+ 2(c^a«)(c6) + 2(a«6*)(ac) + 2(6'c»)(6«).
EXAMPLES. 76
27. {h + cady{hc){ad)+{c + ahdy{c'a)(bd)
'k{a^bcd)*{ab)(cd),
28. Shew that
12{(a; + y + 2p)'"(y + 2;)»"(2P + ajy"(a: + y)'" + a;*' + y** + «**}
is divisible by
(a? + y + «)*  (y + «)*  (« + «)* (a? + y)* + aj* + y* + z\ f^ "^^
29. Shew that
a" (6 + c  a)' + 6» (c + a  6)" + c» (a + 6  c)' + oftc (a" + 6» + c")
+ (a' + 6' + c' ^ 6c  ca  oft) (6 + c  a) (c + a  6) (a + 6  c)
= 2a6c (6c + ca + a6).
80. Shew that
{bcy + {cay + {aby9{bcy{cay(aby
= 2 (a 6)» (a c)« + 2 (6  c)' (6 «)«+ 2 (c  a)» (c  6)'.
81. Shew that
(6 + cy + (c + «)»+ (a + 6)» + (a + rf/+ (6 + rf)' + (c + c^)'
= 3 (a + 6 + c + rf) (a' + 6' + c' + d'),
32. Reduce to its simplest form
4 (a» + a6 + 6»)»  (a  6)« (a + 26)' (2a + 6)'.
83. Shew that
a* (6» + c»  a')' + b' (c" + a»  6^ + c* (a* + 6'  c»)»
is divisible by
a* + 6* + c*  26 V  2c V  2a'6'.
84. Resolve into quadratic factors
4 {cd (a«  6*) + a6 (c'  d')}' + {(a'  6') (c'  e^')  4a6cdf }'. ^ '
85. Shew that
(2/*^){l+xy) (1 +3X5) + {ii?a?) (l + ys)(l + ya;)
+(a;"y*) (l+«B)(l+«y) = (y2) {zx) (xy) (xyzhx^y+z).
86. Find the factors of
a«(6c)(c(^)((^6)6*(crf)(c;a)(ac)
hc^d'a) (a6) {b~d) d^{ab) (6  c) {ca).
87. Find the factors of
6V(^* (6  c) (c  (i) (rf^ 6)  c»(^ V (c d){d a) {a c)
+ dVb' (da)(ab){bd) a'b'c'(a  6) (6  c) (c  a).
CHAPTER VII.
Highest Common FACTORa Lowest Common
Multiples.
95. A Common Factor of two or more integral alge
braical expressions is an integral expression which will
exactly divide each of them.
The Highest Common Factor of two or more integral
expressions is the integral expression of highest dimensions
which will exactly divide each of them.
It is usual to write H.C.F. instead of Highest Common
Factor.
96. The highest common factor of monomial
expressions. The highest common factor of two or more
monomial expressions can be found by inspection.
Thus, to find the highest common factor of a^b^c and a*6^'.
The highest power of a which will divide hoth expressions is
a' ; the highest power of h is h^; and the highest power of c is c ;
and no other letters are common. Hence the h.o.f. is a^l^c.
Again, to find the highest common factor of a^b^t aV)^ and a%(^.
The highest power of a which will divide all three expressions
is a^ ; the highest power of b which will divide them all is & ; and c
will not divide ail the expressions. Hence the h.o.f. is a^&.
From the above examples it will be seen that the
H.C.F. of two or more monomial expressions is the product
of each letter which is common to aU the expressions taken
to the lowest power in which it occurs.
HIGHEST (DOMMON FACTORS. 77
97. Highest common foctor of multinomial
expressions whose fkctors are known. When the
factors of two or more multinomial expressions are known,
their H.c.F. can be at once written down, as in the pre
ceding case. The H.aF. will be the product of ea/chf<wtor
which is com/mon to ail the expressions taken to the lowest
power in which it occttrs.
Thus, to find the h.o.f. of
(« 2)8 (x !)«(« 3) and (a;  2)> («  1) (a?  3)».
It is clear that both ezpressions are diyisible by {x  2)', but by no
higher power of as 2. Also both expressions are divisible by x  1,
bnt by no higher power of a;  1 ; and both expressions are divisible
by x8, bat by no higher power of xZ, Hence the h.o.f. is
(a;2)a(«l)(a!~8).
Again, the h.cf. of a«6»(a6)3(a + 6)8 and a»6*(a6)(a+6)« is
aW(ah)(a+h)\
In the following examples the &ctors can be seen by
inspection, and hence the H.C.F. can be written down,
Ex. 1. Find the h.o.f. of a^lf^a%*^ and a*6»+a»6*.
Ans, a'68(a+6).
Ex. 2. Find tiie h.o.f. of a«6»4a*6* and a^h^l^^lfi,
Ans. a26«(a»46»).
Ex. 3. Find the h.o.f. of a^+Sa^b + 2dli^ and a^+6a>&+8a'&>.
Ans, a (a + 25).
98. Although we cannot, in general, find the feictors
of a multinomial expression of higher degree than the
second [Art. 84], there is no difficulty in finding the
highest common factor of any two multinomial expressions.
The process is analogous to that used in arithmetic to find
the greatest common measure of two numbers.
If the expressions have monomial £Eictors, they can be
seen by inspection; and the highest common factors of
these monomial factors can also be seen by inspection:
we have therefore only to find the multinomial expression
of highest dimensions which is common to the two given
expressions.
78 HIGHEST COMMON FACTORS.
Let A and B stand for the two expressions, which are
supposed to be arranged according to descending powers
of some common letter, and let A be of not higher degree
than B in that letter. Divide B by A, and let the quotient
be Q and the remainder R ; then
B = AQ + R;
.. R = BAQ.
Now an expression is exactly divisible by any other if
each of its terms is so divisible; and therefore B is divisible
by every common factor of A and JB, and R is divisible
by every common factor of A and JS. Hence the common
factors of A and B are exactly the same as the common
factors of ^ and R; and therefore the H.aF. of A and R
is the H.C.F. required.
Now divide A by R, and let the remainder be S ; then
the H.C.F. of R and ^S will similarly be the same as the
H.c.F. of A and R, and will therefore be the H.C.F. re
quired.
And, if this process be continued to any extent, the
H.c.F. 0/ amy divisor cmd the corresponding dividend will
always he the h.c.f. required.
If at any stage there is no remainder, the divisor must
be a factor of the corresponding dividend, and that divisor
is clearly the H.C.F. of itself and the corresponding divi
dend. It must therefore be the H.c.F. required.
It should be remarked that by the nature of division
the remainders are successively of lower and lower dimen
sions ; and hence, unless the division leaves no remainder
at some stage, we must at last come to a remainder which
does not contain the common letter, in which case the
given expressions have no H.c.F. containing that letter.
Since the process we are considering is only to be used
to find the highest common multinomial factor, it is clear
that any of the expressions which occur may be divided or
multiplied by any monomial expression without destroying
the validity of the process ; for the multinomial factors
will not be affected by such multiplication or division.
HIGHEST COMMON FACTORS. 79
Thus the H.c.F. of two expressions can be found by
the following
Rule : — Arrange the two expressions according to
descending powers of some common letter, and divide the
expression which is of the highest degree in the common
letter by the other {if both expressions are of the same
degree it is immaterial which is vsed as the divisor). Take
the remainder, if any, after the first division for a new
divisor, and the former divisor as dividend ; and continue
the process until there is no remainder. The last divisor
will be the H.c.F. required. The process is not used for
fi/nding common monomial factors, these can be seen by
inspection ; a/nd any divisor, dividend, or remainder which
occurs may be multiplied or divided by any monomial ex
pression,
Ex. 1. Find the h.o.f. of as* + a;"  2 and a^ + 2x«  3.
a;»+a«2\aJ» + 2x«3/l
' sfix
2/a; + l
«'+a;2
x^ 1
x^l\a?l(x^l
x1
x1
Thns the h.o.f. is o;  1.
The work would be shortened by noticing that the factors of
the first remainder, namely x'^l, are x1 and x + l. And by
means of Art. 88 it is at onoe seen that x1 is, and that x+1 is
not, a factor of ac* + x*  2.
Ex. 2. Find the h.o.f. of
«' + 4a;^8xy' + 24y' and x^x*y\Sxh/^Sxy*.
The second expression is diyisible by x, which is dearly not a
common factor: we have therefore to find the h.o.f. of the first
expression and a^^afiy + Sxy^Sy*,
s^+^hfQxy^ + 24y^\x* 3^ + Sxu^  8^ (x5y
/ g*+4a:8y8a;V + 24.Ty»^
 5a^y + Saehf^  16«y»  8y*
 5a;8y  20 Af^ + 40^  120 y*
2Qai^^6exy^ + 112y*
80 HIGHEST COMMON FACTORS.
The remainder =28y*(a5'2a:y+4«*): the factor 282/' is rejected
and X*  2xy + 4y' is used as the new divisor.
/ sfi2xhf + i3ey* ^
exhf12xy^+2^y^
6a%~12gy^ + 24y8
Hence jc*  2xy{iy* is the h.o.v. required.
Ex. 8. Find the h.o.f. of
2x*+9aj» + 14aj + 8 and 8«*+ 15a^ + 5a?+10a;+2.
To avoid the inconvenience of fractions, the second expression
is multiplied by 2: this cannot introduce any additional common
factors. The process is generally written down in the following
form :
2aj*+9a:»+14aj+3\3x*+ 16aj»+ 5a^»+ 10a;+2
/2
ex^ + B0afi + 10a^+20x{^(S
6a^+27a^ +42a5+9^
3a^ + 10x^22X'5 \2x*\9sfi+l^ + S
6a;* + 27a:8 + 42a; + 9 /2a;
6a;*420a:»44g«10a?V
7«8+44ar» + 62x + 9
3
21sfi + 132a:» + 166a? + 27/7
21a;8+ 70a?2164a?36^
62 62ar»+310a; + 62
a;2 + 6a; + l\3a:8 + 10aj»22a?6/3a:6
/3ar» + 16a;a+ 8a; ^
ar»+ &x^ 1
 ^x^''2&x^
 6a;8_26a;6
Thus a;' + 5a; +1 is the h.o.f. required.
Detached coeflScients should generally be used [Art. 63].
99. The labour of finding the H.C.P. of two expres
sions is frequently lessened by a modification of the pro
cess, the principle of which depends on the following
Theorem : — The common factor of highest degree in
a particular letter, x suppose, of any two expressions A
and B is the same as the h.c.f. of pA + qB and rA + sB,
HIGHEST COMMON FACTORS. 81
where p, q, r, 8 are any quantities positive or negative which
do not contain x.
To prove this, it is in the first place clear that any
common factor of A and B is also a factor of pA + qB and
oirA^sB.
So also, any common factor of pA 4 qB and rA f sB
is also a factor of s{pA+qB)'q{rA'\sB)y that is, of
{sp — qr) A. Hence, as (^p — qr) does not contain a?, any
common factor of pA + qB and rA + sB must be a factor
of A, provided only that p, q, r, s are not so related that
sp — qr = 0. Similarly any common factor of pA + qB
and rA+sB is also a factor of r (pA + qB) —p (rA + sB)^
that is of {rq —ps) B, and therefore of B.
Since every common factor of A and 5 is a factor of
pA + qB and of rA + sB, and every common factor of
pA + qB and rA+sB ia & factor of A and of jB, it follows
that the H.C.F. of A and B is the same as the H.C.F. of
pA +qB and rA+sB,
Ex. TofindtheH.o.F.of 2a5*+a?6x22x + 3 €knd2x*3a^+2xB.
We have, by snbtraction,
4a!36a;«4a;+6 (I);
and, by addition,
4a^_2a^6x2=2a;«(2a;«a;3) (II).
The required h.cp. is the h.o.p. of (I) and (II), and therefore
of (I) and
2r»a;3 (III).
Multiply (III) by 2 and add (I), and we have another expression,
namely
4a?2x^^=:2x{2x^xB) (IV),
such that the h.cp. of (HI) and (IV) is the h.c.p. required. But the
H.C.P. of (lU) and (IV) is obviously 2sc^xS,
100. If R, S be the successive remainders in the
process of finding the H.c.F. of the two expressions A and
B by the method of Art. 98 ; then, as we have seen, every
common factor of A and JB is a factor of i?, and therefore a
common factor of J. and R. Similarly every common factor
of A and iJ is a common factor of JB and 8, And so on ; so
s. A 6
82 HIGHEST COMMON FACTORS.
that every common factor of A and jB is a factor of every
remainder, and therefore must be a factor of the H.c.F.
Hence every common multinomial factor of two ex
pressions is a factor of their highest common multinomial
factor ; and this is obviously true also of monomial factors.
Therefore every common factor of two expressions is a
factor of their H.C.F.
101. The H.c.F. of three or more multinomial expres
sions can be found as follows.
Let the expressions be A, B, G, D,....
Find O the H.C.F. o{ A and B.
Then, since the required h.c.f. will be a common
factor of A and B, it will be a factor of G : we have there
fore to find the H.C.F. of O, C, D,..,
Hence we first find the H.C.F. of two of the given ex
pressions, and then find the H.C.F. of this result and of
the third expression ; and so on.
Note. The highest common factor of algebraical ex
pressions is sometimes, but very inappropriately, called
their greatest common measv/re (a. CM,).
If one expression is of higher dimensions than another,
in a particular letter, we have no reason to suppose that it
is numerically greater : for example, a' is not necessarily
greater than a; in fact, if a is positive and less than
unity, a' is less than a.
It should also be noticed that if we give particular
numerical values to the letters involved in any two ex
pressions and in their H.C.F., the numerical value of the
H.C.F. is by no means necessarily the G.C.M. of the values
of the expressions. This is not the case even when the
given expressions are integral for the particular values
chosen. For example, the H.C.F. of 14a? + 15a?fl and
22a;' H 23a? + 1 will be found to be a?+l; but if we
suppose a? to be J, the numerical values of the expressions
will be 12 and 18, which have 6 for G.C.M., whereas the
numerical value of the h.cf. will be f .
LOWEST COMMON MULTIPLE. 83
EXAMPLES VII.
Find the H. c. p. of
1. a»  5ab + W and a^  5a'b + 46».
2. 2af6a: + 2 and 12a;' 8aj" 3iB + 2.
3. 2x'3a^y' + y* and 2x''Zxy + y\
4. 2aj' + 3ic*y  2^' and ia^ + xy' ' y\
5. iB'4^"+122/«9;2;' and a:* + 2a32;  4y^ + 8y«  3«'.
6. 20a*  3a'6 + 6* and 64a*  3a6» + 56*.
7. a»  a'b + ab'+l W and 4a' + 3a'6  9ab' + 2h\
8. 2a;* + a'9a:» + 8a;2 and 2aj* 7a;'+ 11a;* 8a; + 2.
9. 11a;*  9aa;»  aV  a* and 13a;*  lOaa;'  2aV  a*.
10. a;* + a;"9a;»3a;+18 and a;* + Ca;"  49a; + 42.
11. aj*2a;« + 5a;».4a;+3 and 2a;*  a;» + 6a;" + 2a; + 3.
12. a;* + 3a;" + 6a; + 35 and a;* + 2a;'  5a;» + 26a; + 21.
Lowest Common Multiple.
102. Definitions. A Common Multiple of two or
more integral expressions, is an expression which is exactly
divisible by each of them.
The Lowest Common Multiple of two or more integral
expressions, is the expression of lowest dimensions which
is exactly divisible by each of them.
Instead of Lowest Common Multiple it is usual to
write L.C.M.
6—2
84 LOWEST COMMON MULTIPLK
103. When the factors of expressions are known,
their L.C.M. can be at once written down.
Consider, for example, the expressions
a^6' (a; — a)' (a; — 6)' and a6* (a? — a)* (a;  6).
It is clear that any common multiple must contain a* as
a factor ; it must also contain b\ (x — a)* and {x — by. Any
common multiple must therefore have a'6* (x — a)* {x — bf
as a factor ; and the common multiple which has no un
necessary factors, that is to say the lowest common multiple,
must therefore be a'6* (x — a)* (x — by.
From the above example it will be seen that the L.C.M.
of two or more expressions which are expressed as the
product of factors of the first degree, is obtained by taking
every different factor which occurs in the expressions to tfie
highest power which it has in any one of them,
Ex. 1. Find the L.0.11. of Sxh/z, 27a?yh^ and 6a;y V.
Ans, 54a%V.
Ex. 2. Find the l.o.m. of Bad^ (a + b)^ and ^a^b (a*  b^).
Ans, 12a^b^{a + by^(ab).
Ex. 3. Find the l.o.m. of 2axy {x  y)^, Saa^ (a?  y^ and 4y2 (x + y)^.
Am. 1200^^2 (x2yy.
Ex. 4. Find the L.O.M. of ix?Zx^2^ x^^ + ^2Ji^ ix?^x¥%,
Ans, {x  1) (05  2) (a;  8).
104. When the factors of the expressions whose L.C.M.
is required cannot be seen by inspection, their H.C.F. must
be found by the method of Art. 98.
Thus, to find the l.o.m. oi a^+a^^ and a:5+2a?3.
The E.G.!*. will be found to be a;  1 ;
and x»+a;«2=(a;l)(a;« + 2«+2),
a;8+2x«3=:(a;l)(x«+3a; + 3).
Then, since x^ + 2a; + 2 and a;' + 3a; + 3 have no common factor, the
required l.c.m. is (x  1) (a;^ + 2x + 2) (a;^ + 3a; + 3).
LOWEST COMMON MULTIPLE. 85
105. Let A and B stand for any two integral ex
pressions, and let H stand for their H.C.F., and L for their
L.C.M.
Let a and b be the quotients when A and B respec
tively are divided hy H; so that
A = H.a and B^H.h.
Since H is the highest common factor of A and B,
a and h can have no common factors. Hence the L.O.M.
of A and B must be if x a x 6. Thus
L=^ H ,a ,h.
Hence L — Hax rf —A x^ (i);
also LxH = HaxHb = A xB (ii).
From (i) we see that the L.C.M of any two expressions
is fofwnd hy dividing one of the expressions by their H.C.F.,
a/nd multiplying the quotient hy the other expression.
From (ii) we see that the product of any two expressions
is equal to the product of their H. c. F. and L. c. M.
EXAMPLES VITL
Find the L.C.M. of
L ea:*  Soaj  6a' and 4a;'  200"  9a'.
2. 4a» bab + V and 3a'  3a'6 + a5»  h\
3. Sa;" 130* + 23a; 21 and 63;" + a;' 44a; + 21.
4. a;*  11a;" + 49 and 7a;*  40a;' + IM  40a; +7.
5. a;' + 6a;' + lla; + 6 anda;* + a;'4a;'4a^
6. a;*  a;* + 8a;  8 and a;* + 43;"  8a;" + 24a;.
7. 8a»  1 8a6", 8a' + 8a"6  6a6" and 4a"  8a6 + 36".
86 EXAMPLES.
8. a:«7fB + 12, 3a;'  6a;  9 and 2aj"  6a;'  8as.
9. 8a;' + 27, 16a;* + 36a;' + 81 and 6a;'  5a;  6.
10. a^exizk 9y% a;'  a;y  6y' and 3a;»  12^.
11. of  7xy + 12^', a;' — 6xy + Sy* and a;' — 5a5y + 6y*.
12. Shew that, if aa^ + bx + c and aV + 6'a; + c' have a com
mon factor of the form x +/, then will
(oc'  a'cy = (6c'  6'c) (a6'  a'b).
13. Shew that^ if ax^ + boi^ + cx + d and aV + b'a^ + c'a; + d'
have a common quadratic factor in x, then will
ba' — 6'a _ ca' — c'a da* — da
okJH — aSd bd' — b'd cd — c'd *
14. Find the condition that aa? + 6a; + c and aV + b'x + c'
may have a common factor of the form x +f.
15. If ff^, g^f ffa are the highest common factors, and Z^, Z^, l^
the lowest common multiples of the three quantities a, 6, c taken
in pairs; prove that 9^g^aW^= {abc)',
16. If A, By G be any three algebraical expressions, and
{BC\ (CA), (AB) and {ABC) be respectively the highest
common factors of B and 0, G and A, A and B, and u4, 5 and
C 'j then the l.o.m. of A, B and (7 will be
A.B.G. (ABG) T {{BG) . {GA) . {AB)},
CHAPTER VIII.
Fractions.
106. When the operation of division is indicated by
placing the dividend over the divisor with a horizontal
line between them, the quotient is called an algebraical
fraction, the dividend and the divisor being called respec
tively the nv/merator and the denominator of the fraction.
Thus 7 means a r 6.
d 1 •. /. 11 ,, . a
Since, by definition, ^ = a ^ 6, it follows that r x 6 = a*
107. Theorem. The value of a fraction is not altered
by multiplying its nuTnerator and denominator by the same
qua/atity.
We have to prove that
a _am
b'^bm*
for all values of a, b and m.
a
Let ^ = r ;
then a?xfe = Tx6 = a, by definition.
88 FRACTIONS.
Hence a?x6xm = axm;
.'. X X (6m) = am, [Art. 29, (ii).]
Divide by 6m, and we have
X = am 7 (6m) = j— ,
108. Since the value of a fraction is not altered by
mvltiplying both the numerator and the denominator by
the same quantity, it follows conversely that the value of
a fraction is not altered by dividing both the numerator
and the denominator by the same quantity.
Hence a fraction may be simplified by the rejection
of any factor which is common to its numerator and
d^x
denominator. For example, the fraction t« takes the
a'
simpler form j^ , when the factor x, which is common to
its numerator and denominator, is rejected.
When the numerator and denominator of a fraction
have no common factors, the fraction is said to be in its
lowest term^.
To reduce a fraction to its lowest terms we must
divide its numerator and denominator by their H.C.F.; for
we thus obtain an equivalent fraction whose numerator
and denominator have no common faotors.
Ex. 1. Bednce , «^ to its lowest terms.
The H.O.F. of the numerator and denominator is Ztjtxy ; and
a'Txy + lOy' _ (a;~2y)(a;5y ) _ xBy
a;28xy + 12y2 (a;2y)(a; 6y) "■ a; 6?/ •
FRACTIONa 89
Ex. 8. Simplify = — ^ .
a*«* ~ (ax){a\x)*
Since a;  a =  (a  x), if we divide the numerator and denominator
by a  «, we have the equivalent fraction ^^— ; and if we divide the
numerator and denominator by xa^ we have — ; r. By the
(a+x) ''
^ £ 2 X
Law of Signs in Division = — ; : = , and the last form
a+x (a + x) avx
is the one in which the result is usually left.
Ex.4. Smiph^ ^^g^^g^^^^^^^ ,
The H.c.F. will be found to be x'>3aB + 7; and, dividing the
numerator and denominator by x*~3a;+7, we have the equivalent
. .. a^ + 3a; + 5
109. Reduction of AractionB to a common de
nominator. Since the value of a fraction is unaltered
by multiplying its numerator and denominator by the
same quantity, any number of fractions can be reduced
to equivalent fractions all of which have the same de
nominator.
The process is as follows. First find the L.C.M. of all
the denominators; then divide the L.C.M. by the denomi
nator of one of the fractions, and multiply the numerator
and denominator of that fraction by the quotient; and
deal in a similar manner with all the other fractions : we
thus obtain new fractions equal to the given fractions re
spectively and all of which have the same denominator.
For example, to reduoe
a h
and
to a common denominator.
The L.O.M. of the denominators is aj^w'Cx'y'). Dividing this
L.c.M. by a^(«+y), xy^{xy) and a:V(*'~y)» ^® ^^® *^^
90 FRACTIONS.
quotients y^{xy), x^{x+y) and xy respeotiTely. Hence the
required fractions are
_ axy^jxy) __ ay*{x,
«V(*+y) ^{x+y)xy^xy) xhf^(x^'y^y
b _ bxa?{x+ y) ba^{x+y)
xy^ixy) xy* (xy) x x^x+y) " a!hf*(a^y^) *
c cxxy exy
«ay2(<pa_ya) ashf^a^  y^fx^ iy^^Zp)*
It is not necessary to take the lowest common multiple of the
denominators, for any common multiple would answer the purpose;
but bj using the L.o.ii. there is some saving of labour.
110. Addition of fractions. The sum (or differ
ence) of two fractions which have the same denominator
is a fraction whose numerator is the sum (or difference) of
their numerators, and which has the common denominator.
This follows from Art. 43.
When two fractions have not the same denominator,
they must first be reduced to equivalent fractions which
have the same denominator : their sum, or difference, will
then be found by taking the sum, or difference, of their
numerators, retaining the common denominator.
When more than two fractions are to be added, or when
there are several fractions some of which are to be added
and the others subtracted, the process is precisely the
same. The fractions must first be reduced to a common
denominator, and then the numerators of the reduced
fractions are added or subtracted as may be required.
Ex. 1. Find the value of — r +
a+b a6'
The ii.o'.v. of the denominators is (a+b) (a > &); and
1 1 _ gft _ a+b
a+b ab (a + b)(ab) {ab){a + b)
(a&) + (a+6) _ 2a
~ a^b* "a^^b^*
a ab
Ex. 2. Find the value of ^—^ + ^_^ •
FRACTIONS. 91
The L.O.M. of the denominators is a^b^; and we have
a{a+b) ah ^a(a'\'b)'ab _ a*
Ex. 8. Simphfy + — —  + ^— ^ + ^— . .
In this case it is not desirable to reduce all the fractions to a
common denominator at once : the work is simplified bj proceeding
asunder:
a a a(a+x) + a(a~x) 2a*
then
ax a+x a*a;* a^a^*
2a« 2a« _ 2a^(a^ + x^+2a^{a*a^) _ 4a*
a*»«"^a«+a;a"" a*a5* "t^ai^'
^ ,, 4a* 4a* 4a*(a*+{B*) + 4a*(a*a;*) Sa^
and finally r — ; + r — ^ = — ^ ^ — y^ ^ = _ — = .
a*x* a*+x^ a^sfi a^ar
[The above would be shortened by observing that, except for the
factor 2, the second addition only differs from the first by having a'
and ae^ instead of a and x respectively; and hence the result of the
addition can be written down at once. So also the result of the
third addition can be written down from the first or second.]
•
Ex. 4. Simplify
053 051 aj + 1 « + 3*
Here again it is best not to reduce all the fractions to a common
denominator at once: much labour is often saved by a judicious
arrangement and grouping of the tenps.
Jl 1 _ (a; + 3)(g~3) _ 6
and
then
3 3 3(g+l) + 3(a;l) ^ 6
6 6 _ 6(a?»l)6(a;«9) 48
a;a9'*"ar»l" (lea  9) («»  1) " (a:*  9) (a;*  1) '
Ex.5. Simplify (^,ftj'(^,e) + (6c) (&a) + (caHc6) '
The L.o.M. of the denominators is {Jbc){ca)(ah) [See Art.
93]. Hence we have
(6c)(ca)(a6)
Now we naturally test, by the method of Art. 88, whether either
of the factors of the denominator is a factor of the numerator: we
are thus led to find that the numerator is the same as the denomi
nator, so that the given expression is equal to unity.
92 FRACTIONS.
Ex. 6. Simplify
a« ^ c*
(a6) (ac) (aj + a) ■*■ (6c) (6a) (a; + &) "^ (<?a) (c 6) (oj+c) •
The L.O.V. of the denominators is
(b^c) ((ja)(a b) (x+a) (x + b) (x+c).
The expression is therefore equal to the fraction whose denominator
is this L.O.M., and whose numerator is
a'(c  b){x + b) {x+c) + h^{a'c) (x+c) {x + a)+e^{ba) («+a) (x+ 6).
Arranging the numerator according to powers of ac, the coefficient
of aj« is aa(c 6) +6*(ac) + c2(6a) = (6c)(c a) (a 6).
The coefficient of x is a* (c^  6*) + 6« (a^  c«) + c^ (fea _ ^2) = 0.
The term which does not contain x is
a6c {a (c  6) + 6 (a  <r) + c (6  a)} = 0.
Hence the numerator is s^{bc){ca){ab)t and therefore the
given expression
x«(&c)(ca)(a&) ^
^{bc)(C'a){ab) (x\a){x + b) (x + c)'~ {x + a) {x + b) {x + c)'
111. Multiplication of fractions. We have now
to shew how to multiply algebraical fractions.
fit G
Let the fractions be t and , .
a
Let ^=6^5'
then a?x6xd = TXjX6x(i
6 a
a , c ,
= Tx6XjXa,
a
by the Commutative Law.
But, by definition, yxb = a, and jxd = c]
_ ??
bd
FRACTIONS. 93
Thus the product of any two fractions is another fraction
whose numerator is the prochict of their numerators, and
whose denominator is the product of their denominators.
The continued product of any number of fractions is
found by the same rule. For
a c e ac e ace
b d f bd f bdf
and similarly, however many fractions there may be.
Hence
/aV a a aa a* , . , /aV a"
Kb) ^b^'b^M = 6 *°^' '"^ general. ^ =^ .
a c
112. Division of ftuctions. Let t and ^ be any
two fractions ; and let a? = r s j .
' b d
mi caeca
Then a?x7 = TTiXj = 7;
d b d d b
c d ^a d
d c b c
Hence » = r x — ,
b c
c d cd ^
smce J X  = 5 = 1.
a oc
Thus to divide by any fraction y is the same as to
multiply by the reciprocal fraction  .
As particular cases of multiplication and division, we
have
a a c _ac
94 FRACTIONS.
J a a c a 1 a
and r^c = r^T = TX = T'
6 1 c be
Note. It should be noticed that the rules for the
multiplication and division of algebraical fractions are
simply rules concerning the order in which certain opera
tions of multiplication and division may be performed, and
have really been proved in Art. 33.
Thus r X ^ = (a "T 6) X (c T d)
= a7b X cTd
= ax cTb7'd = (ac) h (bd) = j^ .
«j3 J. /»3 SR — (L
Ex. 1. Simplify = — r x
x^a^ (x + af'
a^ + a^ xa _ (3ifi + a^)(xa)
(x^ax+a?){x + a) (xa) _ x'*ax\a^
^ (a:a)(a;+a)« " {x{a)^ '
1 1
Ex. 2. Simplify y— 2 .
1 1 y—x
X y xy yx y^x^ yx xhf^ _ xy
1 1 y^ — v? xy x^i/^ ~~ xy y^a^^ x+y'
x^ y5 ~^T
a+x ax
Ex.3. Simplify "^"^ ^'^^ .
a+x ax
+
a—x a+x
d + x __ ax _ {a + x) {a + x){a x) (a  x) __ 4.ax
ax a + x a^x^ "" a«  a;^ '
and ^ + '^ . ^"^ _ {<^ + ^) {a + x) + {ax) {ax) __ 2a^ + 2x'
ax a + x" a^x^ '^~a^x^ '
FRACTIONS. 95
Henoe the given fraction is equal to
4ax . 2a^+2x^ _ 4aas d*x^ _ 2clx
113. The following theorems (the second of which
includes the first) are of importance :
^ Theorem I. If the fractions j^ , j^ , ^ y&cle all
1 s 8
equal to one another , then wUl each fraction he eqiud to
pg^ 4 gg, + ra, +
p\ + qb^ + rb^\
Let each of the equal fractions be equal to x.
Then, since r = a), a.=b.x;
.'. pa^^p\x,
so also qa^ = ^'6, a?,
ra^ = rtg x,
Hence, by addition,
pa^ + qa^ + ra^\ = (pb^ + qb^ + rh^+ ...)x;
pa, + qa^ + ra, +  ^  ^»  ^^e.
p\ + qb^+rb^+ b,
¥ Theorem II. If the fractions ^ , * , ^* , cfec. be all
equal to one another, then will each fraction he equal to
f^ , where A is any homogeneous expression of the nth
degree in a^, a^, ttg, &c. and B is the same homogeneous
expression with 6, in the place of a,, 6, in the place of
a^, (kc.
Let each of the equal fractions be equal to x, so that
Oi = b,Xy a^ = \x, ttg = b^x, &c.
• •
96 FRACTIONS.
Let T^^a/ay... be any term of A; then Xb^^^b/ by...
will be the corresponding term of B ; and since the ex
pressions A and B are homogeneous and of the nth degree,
a + )8 + 7+ ... =w.
Now XOi* a/ ttgY . . . = \ (b^xY (b^xy (b^x)y. . .
since a4)8 + 7+ ... =?*.*
Hence any term of ^ = af x corresponding term of B ;
.*. sum of all the terms of 4 = af* x sum of all the terms of B,
that is A=x* .B\
which proves the theorem.
y Theorem III. If the denominators of the fractions
j^ , j^ , j^ , be all positive, then will the fraction
* 8 8
1 — yi — =^ — '^^^^ be greater than the least and less than
the greatest of the fractions ir y ir* ^^*
Let ^ be the greatest of the fractions, and let ^ = a? ;
then T^<x, T^<x, &a
Hence, fe^, 6,,... being all positive, we have
a^<x. 63,
* We have in the above assumed certain results which wiU be proved
in Chapter XTTT.
FRACTIONS. 97
Hence by addition
ai + a, + a,+ < a? (6^ + 6^ + 63+ );
. a, + a^ + a^\
•• ^ + i. + 6s + ^ '
Hence 7 V 1' . 1" . ' ' is less than the greatest of the
Oj + ftj + ftgH... ^
fractions; and it can be similarly proved to be greater
than the least of the fractions.
Ex.l. Sliewthat,if ^ = .thenwm^± = ^.
6 a a^h ed
€k C
Let r=ip; then =aj.
a
„ a + h _ hx + h x + 1 dx + d c + d
ah~' hX'h" x1^ dzd" cd'
Or thus:
Since f = .
Also Tl= j1, that IS
Hence
b d ' b d
a+b a6__c+d cd ^
a6~cd'
Ex. 2. Shew that, if T^it then will each fraction be equal to
o a
J{b^2bd+2d^)'
Put ^ = =«:;
^^^ V(6^26d + 2d2) V(&a26d + 2d2) V*'^?.
[This is a simple case of Theorem 11., Art. 113.]
Ex.8. Shew that, if ^=^^±^ = ^±^, then mU
bcx __ cay __ 062;
^ al+bm+cn^ albm\cn^ al+bm^cn*
S. A. 7
98 EXAMPLES.
Eaoh of the giyen equal fractions
~a{cy + bz) + h{az\cx) +e{bx ^ay) _ 2bcx
al+bm+cn ~" al+bm{cn
andBimilarly — —
albm + cn cU+bmcn'
EXAMPLES IX.
Simplify the following fractions :
a'  Sab + 7 6' . 7a;V  Safy' + 1
^' a*3a6286«' 2Sxy h Sx'y'  I '
^ (a^y')(a? + y) (a;^  /) (a;  y)
2aj' + 3a;"l ^ a?*iB»a;+l
9.
10.
11.
12.
13.
a;* + 2aj' + 2a»+2a;+r (c*  2aj'  (c"  2a; + 1 '
2a^+5ag'y4a:y*3y'
3ic* + 3a5"y  4a; V  a^ + y* *
54a;'  27a;"  Sa;*  4
363;* + 3aJ» + 3a;  2 *
(a + 6){(af6)'c'}
46V(a*6»cy
a;^ (y ») + y" (««) + «• (a; y)
a?" (ya?) 4y" (ga;) + z^^xy)
(y + «)» + (»+a;)»+(a; + y)»
  g (6  c) (c  c^)  c (c? g) (g  b)
b{c'd){da)''d{a^b){bcy
EXAMPLES. dd
j5 a^(y'~^)Ky'(g'"^)Kg'(a^yO
iB*(y»)+y*(««)+«'(a?y)
,« 2a 26 a* + 6*
16. y + y +
17.
18.
20.
21.
22.
25.
3ag 3 4 a; 1  16 a;
l3a; ITSS "Si?^!*
g y_ _ {x  yy
,^ X'2a x + 2a Scug
19. 5r  Ti +
x+2a 2a'x afia*'
xh2 aj + 4 a; + 6 aj + 8*
JL 3_ 3 1
x+a 05 + 3a a; + 5a a5 + 7a*
1 4 6 4 1
05 — 2a » — a aj os + a a; + 2a'
23. 3.i^:,rlr^.* '
af5xy + Qy' af^+3y* afSxy + 2y''
24. ;^ rr: ^. + r: rr; r +
(ab)(ac) (6c)(6a) (ca)(c6)'
g' 6' c'
(a6)(ac)"^(6c)(6a)"^(ca)(c6)*
(1 f a6) (1 + ac) (l + bc){l+ba) ^ (l+ca)(l+cb)
(a  6) (a  c) (6  c) (6  a) (c a){c b)
^„ bc(a + d) ca{b + d) ab {chd)
(a — 6) (a  c) (6  c) (6  a) (c  a) (c  6) *
(a; + y)(a: + ») (y+«)(y + a) (» + a;)(» + y)'
7—2
lOO EXAMPLES.
„Q (ya?)(^y) . (gy)(^y) .
^^' (oj 2y + z) {x + y2z) (a; + y2«) {2x + y + z)
(zx)(zy)
(2x + y + z) {x 2y + z)'
x + a x^b x + c « (a; + a) (« + 6) (a; + c)
+ r + o ^ ^
xa xb xc (xaUxb)(X'c)
^"' XXX ix^+(bc\'Ca + €^)x
x^a^ xb"^ xc" (aja)(aj6)(ajc)
a« 6» c'
^^ (a6)(ac)'^(6c)(6a)'*'(ca)(c6)
g^ 6^ C^
^^ (a6)(a^c)'^(6c)(6a) + (oa)(c6)'
_, (a + 6)(a + c) , ,, (6 + c)(ft + ^) .  (c + ^X^ + ft)
'■q)<j)°e;)
*\^"c/ \c a/ \a b)
35 1 , 1
(a  6 + c) (a + 6  c) (a + 6  c) ( a + 6 + c)
1
( a^b + c) {ab + cY
bc ca ab
37. Shew that
\x — a x + a af + a ) \x —a /
88. Shew that
a + 6 a6 ^ a*x + b'y .aVftV
ax + by aaiby a'afkb'y' a^x'^b^y*
EXAMPLE& 101
39. Shew that f
(i)
a«
(a  6) (a  c) (1 + ax) {b  4(i.it;» (1 ♦ hx)
c' 1
+
(ii)
(ca)(c6)(l +caj) {I + (ix) (I + bx) (I + ex) '
a h
(ab){a'C){l+ax) {b  c) {b  a) {I + bx)
c —X
(iii)
{c 'a){c b) (l+cx) (1 + ax) (1 + 605) (1+cx)'
1 1
(a  6) (a  c) (1 + aa;) (6  c) (6  a) (1 + bx)
1 aj«
(c  a) (c  6) (1 + ca;) (1 +ax)(l + bx) (l+cx)'
40. Simplify
{a+p)(a + q) ^ {h ^ p) {b ^^ g) ^ {c+p)(chq)
(a  6) (a  c) (a; + a) (6  c) (6  a) (oj + 6) (c—a) {cb) (aj+c)*
41. Simplify
a(b + c — a) 6(c + a — 6) c(a + 6 — c)
+ n :—r; r +
(a  6) (a  c) (6  c) (6  a) (c — a) (c — 6) '
42. Simplify
(a6 + c)(a + ftc) (a + 6  c) ( a + 6 t c)
(a6)(ac) "^ (6c)(6a)
(— a + 6 + c) (a — 6 + c)
(c  a) (c  6)
48. Simplify
a{b + c) 6 (c f g) c (a + ft)
6+c— a c+a— 6 a+b—c'
102 EXAMPLES.
'••.•, •••#*
• •
44!** Sh€f^¥*thaf*rw5ri*— ) +(n + ) +(nm+ — )
• •..•..•.— • V . «K \ nj \ mn)
••••.*•.!.'
(w + — i(w + ){ mn ^ ) = 4.
\ w/ \ nj \ rrm/
45. Shew that
46. Shew that
6 — c c — a a — 6 __ (6 — c)(c — a)(a6)
l+6c 1+ca 1 + a6 " (1 + be) (1 + ca) (1 + oft) '
47. Simplify
/48. Shew that, if
y + «_ z + x _ x + y
b — c c — a a— 6'
then will each fraction be equal to
j{{bcy+{c^ay+{aby}
^ ^49. Shew that, if  = ? , then will
y b'
x+a y+b x+y+a+b
V^50. Shew that, if
X y z
b+c—a c+ab a+b—e*
then will (6  c) a: *• (c  a) y + (a  6) « = 0,
^ EXAMPLES. 103
^61. Shew that, if ^=^Zf£,
and c be not zero, then will each equal ^ — =— ,
^ a — 6
and a(y — «) + 6(«a5) + c(a3 — y) = 0.
52. Shew that
a^ h^ c*
^ J. ^
+ 7:j — m — zTTj — r=a + b\c + a,
(a — a) {db) {d — c)
53. Shew that
a." a'
zi ^ z?
(«i»,)(«,«a) K»J (»«~«i)K»8) K«)
a'
+ ... + «
is equal to zero if r be less than w ~ 1, to 1 if r = n — 1, and to
a^ + a^ + . .. + a^ if r = w.
54. Shew that
ajaj (a;a,)(aja^) (aj  a^) (a;  a.) (aj  a,)
4. «
(ajttj) (a; aj,..{xaj (xa^) (a;a,)...(a;a,) '
55. Shew that
+ 7 TTT 5 X + ...
a(a+6+c+ ... +^ + Q a (a + 6) (a + 6)(a + 6+c)
... +
(a + 6 + ... + A;) (a + 6 + ... + ^ + Z) '
CHAPTER IX.
Equations. One Unknown Quantity.
114. A STATEMENT of the equality of two algebraical
expressions is called an equation; and the two equal
expressions are called the members, or sides, of the
equation.
When the equality is true for all values of the letters
involved the equation is, as we have already said, called
an identity y the name equation being reserved for tjiose
cases in which the equality is only true for certain
particular values of the letters involved.
For the sake of distinction, a quantity which is sup
posed to be known, but which is not expressed by any
particular arithmetical number, is usually represented by
one of the first letters of the alphabet, a, 6, c, &c., and
a quantity which is unknown, and which is to be found,
is usually represented by one of the last letters of the
alphabet x, y, z, &c.
116. We shall in the present chapter only consider
equations which contain owe unknown quantity.
To solve an equation is to find the value or values of
the unknown quantity for which the equation is true ; and
these values of the unknown quantity are said to satisfy
the equation, and are called the roots of the equation.
Two equations are said to be equivalent when they
have the same rpots.
i
L
.^ mi
EQUATIONS. ONE UNKNOWN QUANTITY. 105
An equation which contains only one unknown quantity,
X suppose, and which is rational and integral in x, is said
to be of the first degree when x occurs only in the first
power ; it is said to be of the second degree when a^ is the
highest power of x which occurs ; and so on.
Equations of the first, second and third degrees are
however generally called simple, quadratic and cubic equa
tions respectively.
116. In the solution of equations irequent use is
made of the following principles.
I. An equation is equivalent to that formed by adding
the same quantity to both its members.
For it is clear that J. 4 m = B + ^ when, and only
when, A = B.
IL Any term may be transformed from one side of
an equation to the other, provided its sign be changed.
Let the equation be •
ahb — c^p — q + r.
Add —p + 5 — r to both sides ;
then a + 6 — c— j> + g — r=j> — y + r— jp + g— r,
that is, a + b — c—p + q — r^O,
We thus have an equation equivalent to the given
equation, but with the terms p, —qy +r changed in sign
and transposed.
By means of transposition all the terms of any equa
tion may be written on one side of the sign of equality
and zero on the other side.
IIL An equation is equivalent to that formed by
multiplying (or dividing) each of its members by the same
quantity ^ich is not equal to zero.
For, if J. = 5, it is clear that mA = mB. Conversely, if
mA = mB, that is m ( J.  5) = 0, it follows that AB^O,
since m is not zero. Hence mA = mB when, and only
when, 4=5.
106 EQUATIONS. ONE UNKNOWN QUANTITY.
The case of division requires no separate examination,
for to divide by m is the same as to multiply by — .
117. Simple liquations. The method of solving
simple equations will be seen from the following examples.
Ex. 1. Solve the equation 13a;  7 = Sob + 9.
Transpose the terms 5x and  7; then ISas 5as=7 +9.
That is 8aj=16.
Divide both sides by 8, the coefficient of x; then x=2.
Ex. 2. Solve the equation r  2 = — + 6.
4 5
We may get rid of fractions by multiplying both members by 20,
the least common multiple of the denominators; we then have
16a: 40 = Bar +100,
or transposing 16a; 8a; =100 +40;
/. 7x=140.
Divide by 7» the coefficient of x; then x = 20,
Ex. 3. Solve the equation a{x'a)=z 2ab b{xb),
Bemoving the brackets, we have
ax  a*=2db bx + h\
or transposing tix + hx= 2ab + 6* + a*,
that is x{a+b) = {a+b)K
Divide by a + &, the coefficient of as; then
ja + b)* .
x= z^=a\'b.
a+b
From the above it will be seen that the different steps
in the process of solving a simple equation are as follows.
First clear the equation of fractions, and perform the
algebraical operations which are indicated. Then trans
pose all the terms which contain the unknown quantity
to one side of the equation, and all the other terms to the
other side. Next combine all the terms which contain
tb^ unknown quantity into one term^ and divide by the
SIMPLE EQUATIONS. 107
coefficient of the unknown quantity: this gives the re
quired root
118. Special Cases. Every simple equation is re
ducible to the form ax^b = 0, the solution of which is
x = — . The following are special cases.
I. If 6 = 0, the equation reduces to oo? = ; whence
00 = 0.
II. If 6 = and also a = 0, the equation is clearly
satisfied for ail values of x.
III. If a = 0, and 6 + 0.
Suppose that while 6 remains constant, a takes in
successioD the values ^Kf fjs* fns'"'? ^^^^ will a: take
in succession the values — 106, — 10*6, — 10^6.... Thus as
a becomes continually smaller and smaller, w will become
continually greater and greater in absolute magnitude;
moreover, by making a sufficiently small, x will become
greater tha/n amy assignable quardity ; for example, in
order that the absolute value of x may be greater than
10*** it is only necessary to give to a an absolute value
less than ^^r^ .
10
This is expressed by sayiug that, in the limit, when
a becomes zero, the root of the equation aa? + 6 = is
infinite.
The symbol for infinity is oo .
EXAMPLES.
Solve the eqnations
1. l(a;2)i(aj3) + i(a;4)=4. Am. aj=12.
2. i(«.8)i(aj8)+i(a;6)=0, JLm. ip=0.
108 EQUATIONS. ONE UNKNOWN QUANTITY.
8. a(flfa)=6(a56). Aru, x=a+b,
4. (a5 + a)(a;+6)(aja)(a;6) = (a + 6)'. Aru, a:=^(a+6).
6. a{2za) + b{2xb) = 2ab, Ans, x=^{a\b).
6. (aa + aB)(62 + a;) = (a6+a;)3. Am, «=0.
7. 3(a: + 3)2 + 6(a? + 6)2=8(a; + 8)«. iin«. a:=6.
8. (a:+a)*(aBa)*8aaE?+8a*=0. ^tw. a:=o.
9. («l)8+x» + (a:+l)»=3a;(a;«l). iiw. a;=0.
10. (aj + a)»+(a; + 6)» + (a: + c)8=3(a; + a)(j; + 6)(j; + c).
119. liquations expressed in Factors. It is clear
that a product is zero when one of its factors is zero ; and
it is also clear that a product cannot be zero unless one of
its factors is zero.
Thus {x — 2) (a? — 3) is zero when a? — 2 is zero, or when
a? — 3 is zero, and in no other case.
Hence the equation
(a?2)(a?3) = 0,
is satisfied if a?— 2 = 0, or if a? — 3 = 0; that is, if a? = 2, or
if a? = 3, and in no other case. The roots of the equation
are therefore 2 and 3.
Again, the continued product (a? — »)(«? — 6) (a? — c)...
is zero when a? — a is zero, or when a? — 6 is zero, or when
a?— c is zero, &c. ; and the continued product is not zero
except one of the factors a? — a, a? — 6, a? — c, &c. is zero.
Hence the equation
(aj*a)(a? — 6)(a? — c) ...=0
is equivalent to the system of equations
a?a = 0, a?6 = 0, a?c = 0, &c.
From the above it will be apparent that the solution
QUADRATIC EQUATIONS. 109
of an equation of any degree can be written down at once,
provided the equation is given in the form of a product of
factors of the first degree equated to zero.
Now all the terms of any equation can be transposed
to one side, so that any equation can be written with all
its terms on one side of the sign of equality and zero on
the other side.
^x^It follows therefore that the problem of solving an^
fequ/ati(m of a/ny degree is the sa/me as the problem of finding I
[thefcLctors of cm expression of the so/me degree. ^
Ex. 1. Solve the equation x*  5^; = 6.
Transposing, we have «"  6*  6 = 0,
that is (a;6)(aj+l)=0;
/. a; 6=0, or «+l=0.
Henoe x~6, or x=:l.
Ex. 2. Solve the equation o^  x^ = 6x.
Transposing, we have a:*  a?"  6aj = 0,
that is x(a;8)(9;+2)=0;
/. x=0, or x=3, or x=2.
120. Quadratic liquations. When all the terms
of a quadratic equation are transposed to one side it must
be of the form
• aa? + 6a? + c = 0,
where a, 6, c are supposed to represent known quantities.
We have already [Art. 80] shewn how to resolve a
quadratic expression into factors: the same method will
therefore enable us to find the roots of a quadratic equation.
Hence to solve the quadratic equation
aa? + 6a? + c = 0,
we proceed as follows.
Divide by a, the coeflScient of a?\ the equation then
becomes
a a
112 EQUATIONS. ONE UNKNOWN QUANTITY.
7^
FiXAMPTiKS.
Find the roots of the following equations;
1.
9a^24x + 16=0.
Atu. .
2.
6(a:2 + 4)=4(aj2+9).
i«
^7U. 3^4.
8.
3a;2=8a; + 3.
^fw. 3, 5.
4.
16{B2 + 16a;+3=0.
5.
^+{axY=(a2xf.
^ns. 0, a.
6.
a? + (a2a:)»=(a3a;)a.
Aru. 0, ^.
7.
x^+x^^a^+a.
Ans. a, a1.
8.
x^\'2ax = h'^ + 2db.
^n». b, 2ab.
9.
(xa)«+(x6)2=?(a6)«.
A718, a, b.
10.
(axf+{xhf=.(a^hf.
Am, a, 6.
11.
(6c)aj2+(ca)x+(a6) =
.0.
a'—b
Ans, 1, r .
— c
12.
(x  a+ 26)»  (x  2a+ 6)«=
(a + 6)8.
Afu, a 26, 2a 6.
121. DiscuMion of roots of a quadratic equation.
In the preceding article we found that the quadratic equa
tion cwj* + bx + c = had two roots, namely
"2^"^V 4a' ^^"^ 2a"V~*^~'
Since a / — js — ^^ "^^^ ^^ imaginary according as
V — 4ac is positive or negative, it follows that the roots of
an? + 6a? + c = are real or imaginary according as 6* — 4ac
is positive or negative.
The roots are clearly rational or irrational according
as &' — 4ac is or is not a perfect square. It should be
remarked also that hoth, roots are rational or hoik irrational,
and that both roots are real or hoQi imaginary.
SPECIAL FORMS OP QUADRATIC EQUATIONS. 113
If 6* — 4ac = 0, both roots reduce to — ^ , and are thus
equal to one another. In this case we do not say that
the equation has only one root, but that it has two equal
roots.
It is clear that the roots will be unequal unless
6* — 4ac = 0. Hence in order that the two roots of the
equation aa? + 6a? + c = may be equal, it is necessary and
sufficient that V = 4ac.
When V = 4ac, the expression ax* + hx + c is a perfect
square in x, as we have already seen. [Art. 83.]
122. Special Forms. We will now consider some
special forms of quadratic equations, in which one or more
of the coefficients vanish.
I. If c = 0, the equation reduces to
aa? + 6a? = 0,
or X (ax + 6) = 0,
the roots of which are and — .
a
II. If c = and also 6 = 0, the equation reduces to
aa^ = 0, both roots of which are zero.
III. If 6 = 0, the equation reduces to ax* + c = 0, the
roots of which are ± ^ / — . The roots are therefore
eqital and opposite when 6 = 0, that is when the coefficient
of X is zero.
IV. If a, 6 and c are all zero, the equation is clearly
satisfied for all values of x,
V. If a and 6 be zero but c not zero,
put a? =  in the equation cw?" + 6a; + c = ;
if
S. A. 8
114 EQUATIONS. ONE UNKNOWN QUANTITY.
then we have, after multiplying by y",
cy* + 6y + a = 0.
Now from I. and II. one root of this quadratic in y
is zero if a = 0, and both roots are zero if a = and also
6 = 0.
But since a? = , a? is infinity [Art. 118] when y is zero.
Thus one root of aa? + 6aj + c = is infinite if a = ; also
both roots are infinite if a = and also 6 = 0.
Thus the qoadratio equation
(a  aO rr' + (b  6') a; +c  c'=:0
has one root irdinite, if a^a'\ it has two roots infinite^ if a:=a' and
also 6=5'; and the equation is satisfied for all values of x, if a=a\
5=6' and c=c'.
Again, the equation
a(« + 6)(a? + c) + 6(«+c){«+o)=c(a; + a)(a;+6),
is a quadratic equation for all values of c except only when c=a\l>^
in which case the coefficient of x^ in the quadratic equation is zero.
When c=a+5 we may still however consider that the equation
is a qiuidratic equation, but with one of its roots inAnite,
Note. It is however to be remarked that since in
finite roots are not often of practical importance in Algebra,
they are generally neglected unless specially required.
123. Zero and infinite roots of any equatibn.
The most general form of the equation of the nth degree is
CMf \ bx''~^ +...+ kx + I =^ (i).
If Z = 0, the equation may be written
one root of which is clearly zero.
Similarly two roots will be zero it 1=0 and also A? = ;
and 80 on, if more of the coefficients from the end vanish.
EQUATIONS NOT INTEGRAf.. 115
Put a! = i; then we have, after multiplying by y",
a + by + + A:y*"' + iy" = 0.
From the above, one root of the equation in y will be
zero when a = 0; and two roots will be zero if a = and
also 6 = 0. But when y = 0, » =  = oo.
y
Thus one root of (i) is infinite when a^O, and two
roots are infinite when a and b are both zero ; and so on,
if more of the coefficients from the beginning vanish.
124. Equatione not integral. When an equation
is not integral, the first step to be taken is to reduce it to
an equivalent integral equation.
An equation will be reduced to an integral form by
multiplying by any common multiple of the denominators
of the fractions which it contains, but the legitimacy of
this multiplication requires examination. For if we
multiply both sides of an integral equation by an ex
pression which contains the unknown quantity, the new
equation will not only be satisfied by all the values of the
unknown quantity which satisfy the original equation, but
also by those values which make the expression by which
we have multiplied vanish. Thus if each member of the
equation J. = iS, be multiplied by P, the resulting equa
tion PA = PB, or P (J. — B) = 0, will have the same roots
as the equation J. — iS = together with the roots of the
equation P = 0.
however an equation contains fractions in w];u)0e
denominaEdrs^^^e unknown quantity occurs, tJ)^»^(}uation
may be multipliechby^^ lowest commoimiultiple of the
denominators without inlit)duQingjMi^additional roots, for
we cannot divide both side§uef ^^Eh^oc^lting equation by
any one of the factor^jrf'me L.C.M. wiQlOHtjceintroducing
fractions, whicJi.*4Bhews that there are no roots ^6fHhfi re
6ultingeq«r{C6[on which correspond to the factors of th&
lion by which we multiply.
82
116 EQUATIONS. ONE UNKNOWN QUANTITY.
8 2x
Ex. 1. Solve the equation = + = 6.
•C ~" O iC ~" »J
Multiply by {x5) (as 3), the l. o. m. of the denominators; then
we have
3(a:3) + 2a5(a;6)=6(aj6)(aj8);
/. Sar' 33a; + 84=0.
Whence a:=4 or x=7,
a;*3a; 1
Ex. 2. Solve the equation —5 — ^ + 2 + 7 = 0.
X*— 1 x — l
Multiply by x^  1, the l. 0. m. of the denominators ; then we have
a;«3x + 2(xal)+a;+l=0,
which reduces to
3a;22a?l = 0,
that is (3a}+l)(a:l)=0.
Thus the roots appear to be  ^ and 1; the latter root is however
due to the multiplication by x^  1.
Q. sfiSx 1 _ a;'3a; + x + l _ (acl)* a;l
the equation is equivalent to
xl
42=0,
x+1
which has only one root, namely a;=  J.
From the above example it will be seen that when an equation
has been made integral by multiplication, some of the roots of
the resulting equation may have to be rejected.
Ex. 3. Solve the equation :
x g9 _ x+l xS
«2"*"x^"«l 056'
In this case it is best not to multiply at once by the l.o.m. of the
denominators of the fractions; much labour is often saved by a
judicious arrangement and grouping of the terms.
By transposition we have
X as+l x9 «8
082 081 a;7 086
The first two terms = 7 ^, =, ,
(x2)(xl)'
=0.
EQUATIONS NOT INTEQBAL. 117
2
and the other terms =
(a;7)(a?6)'
Henoe the equation is equivalent to
2 2
=0.
(aj2)(«l) {x7){x'6)
Now mnltiplj by the l.c.h. of the denominators; then
2(aj7)(a;6)2(«2)(a;l)=0,
whioh rednoes to
20«80=0;
Or thus:—
The equation is equivalent to
' _i +£z9_i =£±»_i +£r8_i,
x2 x1 x1 a:6
2 2 2 2
that is — s 5 = — 7 s ♦
a;2 aj7 a;l a;6
10 10
(«2)(aj7)"(a?l)(a?6)'
from which we find as before that x=:4k,
Ex. 4. Solve the equation :
a b c n
aj + a a;+6 aj+c
Wehave ^1 +^1 +^1=0;
x+a x+b x+c
«+a «+6 x + e
«
Hence «=0 (i)»
or else — :— + — ? + — t" =^'
a:+a aj+6 a?+c
Multiply by the L. O.K.; then
{x\b){x+c) + (x{c){x+a)^{x+a)(x+b)=0,
that is
8a^+2a;(a+6+c) + 6c+ca+a6ssO,
118 EQUATIONS. ONE UNKNOWN QUANTITY,
the roots of whioh are
{(a+6+c)±^/(a«^62 + c»6ccaa^>)} (ii).
Thus there are three roots given by (i) and (ii).
Ex. 6. Solve the equation:
h+c e + a a+6 a+h+c
+ +
bcx eax dbx
The equation is equivalent to
6 + c a c + a h a+6 c ^
\ — I —  — — =0,
be^x X cax x abx x
Taking the terms in pairs we have
(a + 6 + c)aJatc {a + b+c)xahe {ahb + c)xahc ^
X (be x) 05 (ca  «) x(abx) ~ '
Hence (a+6 + c)a:a6c=0 1,
or — rr r +7 x + — TT x=0 H.
x{f>cx) x(ca—x) x((Wx)
From I. we have «=
a{'b\c'
From n. we have on multiplication by the l.c.m.
(ca  a;) (oft a?) + {ab o^ipcx)\ibc x) {ca  ac) =0,
that is 8058  2a5 (6c + ca +a6)+o6<j (a + 6 +c)=0,
whence a8= J {6c + ca+a6±^6V+c2a*+a26*a6c (a+6 + c)}.
125. Irrational Equations. An irrational equa
tion is one in which squaxe or other roots of expressions
containing the unknown quantity occur.
In order to rationalize an equation it is first written
with one of the irrational terms standing by itself on one
side of the sign of equality : both sides are then raised
to the lowest power necessary to rationalize the isolated
term; and the process is repeated as often as may be
necessary.
IRRATIONAL EQUATIONS. 119
Ex. 1. Solv* the equation Jx+i + iJx + 20  2 Jx+H = 0.
We have Jx + i+Jx+20=:2 Jx+ll,
Square both members : theu
2a? + 24 + 2 ^/STi ^ic + 20 = 4 (a; + 11),
which is equivalent to
Jx+l iv/iB+20sBaj+10.
Square both members : then
(a; + 4)(«+20) = (a;+10)2,
whence a; =5.
Ex. 2. Solve the equation x/2x + 82^a; + 5=2.
Square both members: then
2a;+8 + 4(a;+5)4/s/2«+8,y«T5=4;
.*. 3a? + 12 = 2 j2x + 8 Jx+5.
Square both members : then
9x3+72a?+144=4 (2a;+8) (ar+6);
.. a?«=16,
whence a?=4 or aj=4.
Ex. 8. Solve the equation ^ax{ a\' tjbx+p+jcx {ys^Oi
We have ,Jax + a\ Jbx+p^ Jcx+y.
Square both members: then we have after transposition
{a+bc)x + a+py= ^2»Jax + a jjbx+p.
Squaring again, we have
{(a+6c)a5 + a + j87}»=4(aa? + a)(6a?+/3),
that is aja {a^+b^ + c^2bc  2ca2db)
+ 2x (aa + bp + cy by  cp  ca  ay  op ^ba)
+ a2+/3»+7«  2/S7 27a  2a/3=0.
Thus the given equation is equivalent to a quadratic equation.
It should be observed that it is quite immaterial what sign
is put before a radical in the above examples; for there are two
square roots of every algebraical expression and we have no symbol
120 EQUATIONS. ONE UNKNOWN QUANTITY.
wh ich re presents o ne on ly to the exdiunon of the oth er; s o that
+ Jx \l and ijx+l are alike equiva lent to ^,J»+1; also
x+ tJx + 1 has the same two yalues as x^^Jx+l,
126. By squaiing both members of the rational equa
tion A=B, we obtain the equation J.* = 5' ; and the
equation J.' = 5*, or J.' — jB* = 0, is not only satisfied when
J. — B = 0, but also when A{B = 0, Hence an equation
is not in general equivalent to that obtained by squaring
both its members; for the latter equation has the same
roots as the original equation together with other roots
which are not roots of the original equation. Additional
roots are not however always introduced by squaring both
sides of an irrational equation. For example, the equation
a? + 1 = ^/a?+13 is really two equations since the radical
may have either of two values ; and by squaring both
members we obtain the equation (a? 41)* = a? +13, which
is equivalent to the two. [See Art. 162.]
127. A quadratic equation can only have two
roots. We have already proved that an expression of the
nth degree in x cannot vanish for more than n values of a?,
unless it vanishes for all values of x. This shews that an
equation of the nth degree cannot have more than n roots,
and in particular that a quadratic equation cannot have
more than two roots.
The following is another proof that a quadratic equa
tion can only have two roots.
We have to prove that aof ^hx + c cannot vanish for
a, p, 7 three unequal values of x. That is we have to
prove that
oaf +6a +0 = (i),
a^ + 6/3 + c=0 : (ii),
and 07' +67 + = (iii),
cannot be simultaneously true, unless a, 6, c are all zero.
A QUADRATIC EQUATION HAS ONLY TWO ROOTS. 121
From (i) and (ii) we have by subtraction
a(a«^^) + 6(a/3) = 0,
that ia (aiS) {a(a + l3) + b} = 0.
But a— ^ + 0; hence
a(a+i8) + 6 = (iv).
Similarly, since ^ — 7 + 0, we have from (ii) and (iii)
a(l3 + y) + b = (v).
From (iv) and (v) we have by subtraction
a (a — 7)5=0 (vi).
Now (vi) cannot be true unless a = 0, for a — 7 ^ 0.
Also when a = 0, it follows from (iv) that 6 = 0, and then
from (i) that c = 0.
Thus the quadratic equation a^ + 6^ + c = cannot
have more than two different roots, v/nless a = 6 = c = ;
and when a, by c are all zero it is clear that the equation
do? + 6aj + c = will be satisfied for all values of a?, that is
to say the equation is an identity,
Ex. 1. Solve the equation a' ^^'^^ l^'^l + d^ t^^'lt^^'U ^'
The equation is clearly satisfied by x=a, and also by as= b; hence
a, h are roots of the equation, and these are the only roots of the
quadratic equation. [The equation is not an identity, for it is not
satisfied by a;=c.]
Ex. 2. Solve the equation
, (a;~5)(x~c) j^i (xc){xa) Jxajix^b)
(ah)(a'cy '{bc)(ha)^ (ca){cb)
The equation is satisfied by »=a, by a;=3&, or by a;=c. Hence,
as it is only of the second degree in x, it must be an identity,
Ex. 3, Solve the equation
{xb){X'C ) (x^c)(X'a) a(a?a)(a;6)_ ,
(a6)(ac)^ (b'C)(bay {c'a){cb)''^'
122 EQUATIONa ONE UNKNOWN QUANTITY.
The equation is satisfied by xssa, by x=h, and by «=c ; and the
equation is not an identity, since the coefficient of x* is not zero.
Hence the roots of the cubic are a, b, c*
Ex. 4. Shew that, if
(a  a)»x + (a  j8)«y + (a  y)*z = (a  «)»,
(6  a)«ar+ (6  /3)« y + {b yfz = (6  3)a,
(co)>a;+(ci8)«y + (c7)«« = (c«)«,
then will
(d  a)«ar + (d  /3)«y + (d  7)«« = (d  3)2,
where d has any value whatever.
The equation
(Xa)>a5+(Z/3)»y + (Z7)'£; = (X5)3
is a quadratic equation in Z, and it has the three roots a, &, e. It is
therefore satisfied when any other quantity d is put for X,
128. Relations between the roots and the coeffl
oients of a quadratic equation.
If we put a and ^ for the roots of the equation
aa?'\hx + c= 0, we have
6 /6*4ac
and
^^^y
4a'
By addition we have
a+/9 =  (i).
By multiplication we have
. «^ = i^« 4^=a ^">
The formulae (i) and (ii) giving the sum and the
product of the roots of a quadratic equation in terms of the
coefficients are very important.
RELATIONS BETWEEN ROOTS AND COEFFICIBNTS. 123
129. Relations between the roots and the oo
efficients of any equation. By the following method
relations between the roots and the coeflBcients of an
equation of any degree may be obtained.
We have seen that if the expression of the nth degree
in X
aoi" + hoT^ + onT^ + cfe^ + ... ,
vanish for the n values a?=a, a? = /8, a?=: 7, &c., then will
cw^ + 6a;""* + car* + (ic""» + ...=:a(a?.a)(aj/8)(a?7)...
We have therefore only to find* the continued product
(a? — a) (aj — iS) (a? — 7) and equate the coeflScients of the
corresponding powers of x on the two sides of the last
equation.
For example, if a, ^, 7 be the roots of the cubic equa
tion oa?" + feic' + CO? + d = 0, we have
cw^ + 6a;* + ca? + d = a (a? — a) (a? — iS) (a? — 7)
= a {a;*  (a + ^ + 7) a?" + (^7 + 7a + ai8) a? 0^7}.
Hence, equating coefficients, we have
6 1
a + /8+7 = .
. a
^7 + 7a + a^= ^1
P d
afiy =  
a
It should be remarked that the sum of the roots of any
equation will be zero provided that the term one degree
lower than the highest is absent*.
We may make use of the above to prove certain identical rela
tions between three quantities whose snm is zero. For a, b, c will
be the roots of the cubic «* ^px+q = 0, provided that a + 6 + c = 0, and
that p and q satisfy the relations
* See Art. 260,
124 EQUATIONS ONE UNKNOWN QUANTirY.
be+ca + ah=p (i),
dbc= q (ii).
Then, sinoe a+d+c=0 (iii),
we have a'+6*+c'=(a+6 + c)'2(6c+ca+a&)
= 2p (iv).
Also, sinoe a, h, c are roots of x^+px + 9=0,
a'+)a+g=0 J
6»+jp6+ff=0 I (v).
c*+>c+g=oJ
From (y) by addition
a»+6»+c»=8g (vi).
Multiply the equations (v) in order by a*"', 6""^, c*~', and add;
then
a* + 6* + c* +!> (a*~* + 6""* + c**) + q (a*"' + fr*""' + c*») = 0.
Hence we have in succession
a* + M+c*=2p',
Hence also
a»<y+cg _ a'+y+c» a»+y+c>
6 2 • 8 •
7 " 2 • 6 •
^ 3 4
• 1 •
[See also Art. 308, Ex. 2.]
130. Equatione with given roots. Although we
cannot in all cases find the roots of a given equation, it is
very easy to solve the converse problem, namely the
problem of finding an equation which has given roots.
For example, to find the equation whose roots are 4 and 5.
We want to find an equation which is satisfied when as=4, or
when xs=5; that is when jc ~ 4=e0, or when x  5=0; and in no other
cases. The equation required must be
(a;4)(a;6) = 0/
that is, ««  9«+ 20=0,
EQUATIONS WITH GIVEN BOOTS. 12&
for this is an equation which is a tme statement when a; 4=0, or
when a;  5=0, and in no other case*.
Again, to find the equation whose roots are 2, 8, and  4.
We have to find an equation which is satisfied when a; 2=0, or
when a;  3 = 0, or when x + 4 = 0, and in no other case. The equation
must therefore be {x2)(x^ 8) (« + 4) = 0,
that 18 x^x*14x^2i=0.
Ex. 1. If a, j9 are the roots of the equation iu^+bx{c=0, find the
equation whose roots are — and  .
P «
The required equation is
(?)(!)»•
that is «*ar^5i^ + l=0.
Now, by Art. 128, we have
5 c
a+^=, ap=;
..a tp ^, ^^,
aj8 "" \a* a) ' a^
2ae
ae
Hence the required equation is
, b^2ac  ^
aa a? + l=0.
ac
Ex. 2., If a, j3, 7 be the roots of the equation aa^ + bx?+cx+d=0,
find the equation whose roots are py^ ya, aj8.
The required equation is
{xPy){xya)(xap)=0,
that is a»«'(/37 + 7a+aj8) + aa/S7(o+i8+7)a«/3V=0.
• The equation a:*  9x + 20 = is certainly an equation with the
proposed and with no other roots ; but to prove that it is the only equa
tion with the proposed and with no other roots, it must be assumed that
every equation ha» a root,
U, for example, the equation a:'^+7a:'2 = had no roots, then
{x  4){x  6) (x^+ 7n^  2) =0 would also be an equation with the proposed
roots and with no others.
The proposition that every equation has a root is by no means easy to
prove; the proof is given in works on the Theory of Equations.
126 EQUATIONS. ONE UNKNOWN QUANTITY.
Now, by Art. 129, we have
a+i8 + 7=,
Py+ya + a^^,
and
a^7=^
Heno
e the required equation is
. c ^^ db d2
a a^ or
or a^afiaca^+bdxcP=0,
131. Changes in value of a trinomial expression.
• The expression cux^ + 6aj + c will alter in value as the value
of X is changed ; but, by giving to x any real value
between — oo and + oo , we cannot make the expression
aa? + bx + c assume any value we please.
We can find the possible values of aa? + bx + c, for
real values of w, as follows.
In order that the expression aa? + bx + c may be equal
to \ for some real value of x, it is necessary and sufficient
that the roots of the equation
oof + 6aj + c = \
be real, the condition for which is
6«_4a(cX)>0,
that is 6'— 4ac + 4aX>0 (i).
I. If 6' — 4ac be positive, the condition (i) is satisfied
for all positive values of 4a\, and also for all negative
values of 4a\ which are not greater than 6' — 4ac.
Thus, when 6* — 4ac is positive, oaf + 6a? + c can, by
giving a suitable value to a?, be made equal to any quantity
of the same sign as a, or to any quantity not absolutely
greater than — j and whose sign is opposite to that of a.
n. K 6* — 4ac be negative, the condition (i) can
only be satisfied when 4ta\ is positive and i^pt less than
CHANGES IN VALUE OF A TRINOMIAL EXPRESSION. 127
Thus, when b^^^ac is negative, ax^+bx + c must al
ways have the same sign as a, and its absolute magnitude
can never be less than — ,
4a
III. If 6* — 4ac be zero, the condition (i) is satisfied
for all positive values of aX.
It follows from the above that the expression aa?+ Ix+c
will keep its sign unchanged, whatever real value be given
to w, provided that 6* — 4ac be negative or zero, that is
provided that the roots of the equation ao? + 6a? + c = be
imaginary or equal, and also that the expression can be
made to change its sign when the roots of ao? + ia? + c =
are real and unequal. We give another proof of this
proposition.
If the equation oaf + 6aj + c = have real roots, a, /8
suppose, then oo?' + 6aJ + c = a (a? — a) (a:  ^).
Now (a? — a)(a; — ^) is positive when x has any real
value greater than both a and ^, or less than both a and
B ; but (x — a) (ic — ^) is negative when x has any real
value intermediate to a and ^.
Thus for real values of x the expression aa^+bx+c
has always the same sign as a except for values of x which
lie between the roots of the corresponding eqtiati<m
aa? + 6a? h c = 0.
132. We can also prove that the expression aaj^+6a?4c
will or will not change sign for different values of x accord
ing as 6' — 4ac is positive or negative, as follows.
I. Let 6* — 4ac be positive.
The whole expression within square brackets will
clearly be negative when a? = — 5 ; also, when x is very
128 EQUATIONS. ONE UNKNOWN QUANTITY.
great, (^ + 9 ) will be greater than — , , and there
fore the whole expression within square brackets will be
positive.
Thus when V ^4^(10 is positive the expression cux^^boD+c
can be made to change its sign by giving suitable real
values to a,
II. Let 6*— 4ac be negative (or zero).
Since (a? + 9") is positive for all real values of x, and
75 — ^s also positive (or zero), the whole expression
within square brackets must be always positive.
Thus when 6* — 4ac is negative or zero, the expression
aa? + bx + c will always have the same sign as a.
133. It follows from Article 131 or 132 that if an
expression of the second degree in x can be made to
change its sign by giving real values to x, then must the
roots of the corresponding equation be real.
Consider, for example, the expression
a' (a?  /8) (a?  7) + 6* (x  7) (a?  a) + c' (x  a) (a?  /5),
where the quantities are all real, and a, 13, 7 are supposed
to be in order of magnitude. The expression is clearly
positive if 4? = a, and is negative if a? = ^. Hence the
expression can be made to change its sign, and therefore
the roots of the equation
a'(a?/8)(a?7)+6«(a?7)(a?a) + c'(a?a)(a;/8) =
are real for all real values of a, 6, c, a, /8, 7.
Ex. 1. Shew that (a;l)(«3)(a;4)(a;6) + 10 is positive for aU
real values of as.
Taking the first and last factors together, and also the other two,
the given expression becomes
(«»7a? + 6) (x«7a?+12) + 10
»(«»7i?)«+18(aJ»7«) + 82
= {(a;'7x) + 9p + l,
which is clearly always positive for real values of «•
EXAMPLES. 129
Ex. 2. Shew that, by giving an appropriate real value to or,
4^+ 86^ + 9
=Ts — 5 7 can be made to assume any real value.
^ . 4a;'+36«+9
^"* Ux^+Sx + l'^*
then »«(412X) + (368X)a;+9X=0.
Now in order that x may be real it is necessary and sufficient
that
(36  8X)«  4 (4  12X) (9  X) >0,
or that X28X + 72>0,
or (X4)2+56>0,
which is clearly true for all real values of X. Thus we can find
real values of x corresponding to any real value whatever of X.
/pQ _ 3a; + 4
Ex. 3, Shew that s — ^ ^ can never be greater than 7 nor less
ar+3x + 4
than I for real values of x.
«_^ x^Sx + i
Put 5 — 5 r=X;
then a?2(lX)3a;(l + X)+4(lX)=0.
In order that x may be real it is necessary and sufficient that
9(1+X)«16(1X)2>0,
thatis 7Xa+60X7>0,
or _(7Xl)(X7)>0.
Hence 7X  1 and X  7 must be of different signs, and therefore
X must lie between = and 7, which proves the proposition.
EXAMPLES X.
Solve the following equations :
1. (a;a + 26)»(a:2a + 5)» = (a + 5)».
V 2. (c + a26)aj' + (a + 62c)a; + (6»c2a) = 0.
{xay {x + hY'
M a + x b + x „,
6 + a? a + x ^
S. A. 9 /
130 EQUATIONS. ONE UNKNOWN QUANTITY.
<zx+b cxAd
5.
6.
11.
19.
a + hx c + dx'
a — 03 _\bx
1 oo; h — x '
7. = = a^+2x
X+l 05+1 '
a* x' 5a;  4
8. a; + 1 +
05* 1 aj + 1 a:*! *
11 1 1a
058 056 X + x+o
,n 2 5 3 4
10. x+ ;; =
a;+8 aj+9 a3+15 a; + 6*
2 . 1 6
2a;3 "^ a;2 ~ 3a;+ 2*
,rt x—a 055 050* «
12. T + + = 3.
036 xc xa
^ a; + a a; + 6 x + c «
13. + ^ + = 3.
a — 03 6 — 03 C — 05
14. + . + = 3.
05 — a 036 03 —
. 2a: 1 3a; 1 , x7
16. — = + = 4 + > .
a;+l a;+2 xl
^^ X 2 X Z
16. +=+.
2 a; 3 a;
_ x + a 05a 03 + 6 03 — 6 ^
17. + + +  = 0.
05 a 03 + a 036 05 + 6
,Q a;l a;4 a;2 a;3
lo. = +
a; + l a;+4 a; + 2 a;+3*
1 1
1 T"'
X'\a\ — — i X"a{
03 + 6 036
EXAMPLES.
131
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
^35.
1 1 1
(^^h a + c _ 2 (a + 6 + c)
aj + 6 x + c "" x + h + c *
^ + c 6t>c _ a + 6 + 2c
35 + 25 a; + 2a "" x + a\h *
a;6 a?a ' 2 (a 5)
{x + a){x{h) ^ {x{c){x + d)
x+a\h x\c + d
a(p;d) _^ d(ab) ^ b(cd) c(a h)
x + a x + d x + b ^
x + e
x — a 03 — 6
a
b
ab
a
05 — a X b'
b^
c^a
= 0.
05 + a — 6 x + bc xic — a
12 3 4
1 + 2aj 2 + 3aj 3 + 4a; 4 + 5aj
(a?  g) (a;  6) (a; f a) (a; + 6)
(ajma;(a;7w5) " (a; + ma) (a; + rw^)) *
V2a; + 97a^^=^/^TT.
V(ajl)(a;2) + V(aj3)(aj4) = V2.
V7aj 5 + V4a;1 = V7a;4 + V4a; 2.
Va*  aj + Vft" + a; = a + 6.
= 0.
's/a'bx + i^cdx = \/a'^c{b'^d)x.
36. ^axhb' + V6aj + a' = a 6.
9—2
132 EQUATIONS. ONE UNKNOWN QUANTITY.
37. Ja + x+Jb\'X==Ja + b\ 2x.
38. Jax + JbT~x = J2a + 26.
39. J{a + x){x + b) + J(ax){xb) = 2 7^.
40. Ja{a + b + x)  Ja{a + bx) = a;.
41. Jaf + ax + b'''Jix^aa:¥b'=^2a.
42. Jaf + ax + a' + Jaf  ax + a'' = J2a'  2b'.
43. Jax^kJ^iiTb = Jax + b + Jcxb.
44. Jx(a + bx) + Ja{b\xa) + 7^ (a + a;  6) = 0.
45. Jx + a + Jx'\b + Jx + c = 0,
46. Jab{a + b + x) = Ja{a¥b){b x) + Jb{a + b)(ax).
47. Jx' b' <f + Jaf c' a' + Jaf  a' b^ = x.
 48. J^n^+JV^^+J7^^=Ja' + b'^c'o^.
^49. For what values of aj is ^14 (3a;2)(ajl) real.
a*+ 34a; — 71
' ;. ' 50. Shew that— "5 — ^r = can have no real value between
ar + 2a; — 7
5 and 9.
51. Shew that, if a; be real 5 — . = can never be less
' ar + 2a; + 1
than — ^.
JK* _ Qf* ^ X
52. What values are possible for ^ — ^ , a; being real.
X ^ X "r X
53. Find the greatest and least real values of x and y
which satisfy the equation a;* + ^ = 6a; — By.
54. Find the greatest and least real values of x and y when
aj« + 4y"8a;16y4 = 0.
55. When x and y are taken so as to satisfy the equation
(a;* + y*y  2a* (a;* — y^\ find the greatest possible value of y.
EQUATIONS OF HIGHER DEGREE THAN THE SECOND. 133
56. . Shew that if the roots of the equation
be real, they will be equal.
57. If the roots of the equation aa^ + 6a; + c = be in the
ratio m : n, then will mnb' = (m + n)' ac,
58. If CKc" + 2hx + c = and a'a? + 2h'x + c'= have one and
only one root in common, prove that b' — ac and 6'" — a'c' must
both be perfect squares.
59. If a5i, 03 be the roots of the equation ooj* + ftoj + c = 0,
X* x'
find the equation whose roots are (i) a^* and a?,*, (ii) J and ^'
(iii) b + ofiCj and b + aac^.
OJ, X,
60. If ajj, a5j be the roots of ckc* + fta; + c = 0, find in terms
of a, 6, c the values of
x^ (bx^ + c) + a?/ (6a;j + c), and aj^' (ftaj^ + c)' + aj^,' (ftaj^ + c)".
61. Shew that, if aj, , ai^ be the roots of aj* + mx + m* + a = 0,
then will x^' + aj^aj^ + a;,* + a = 0.
62. If ajj , a!g be the roots of (»* + 1) (a' + 1) = wmmc (oa?  1),
then will (aj," + 1) (a?,* + 1) = majja;^ {XiX^ — 1).
63. If ajj, a;, be the roots of the equation
A{ac^ + w') + Amx + ^wV = 0,
then will A (x^^ + a;,') + iajja;, + Bx^\^ = 0.
64. Prove that, if a; be real, 2{a — x) (x + Jx' + b^) cannot
exceed a* + b',
\/ 65. Find the least possible value of j^ — y^ ,
for real values of o^
134 EQUATIONS. ONE UNKNOWN QUANTITY.
Equations of higher degree than the second.
134. We now consider some special forms of equations
of higher degree than the second, the solution of the most
general forms of such equations being beyond our range.
V 135. Equations of the same form as quadratic
equations.
The equation cm?* + 6a?' + c =
can be solved in exactly the same way as the quadratic
equation ao? + 6a? + c = ;
we therefore have
, ^ h JV 4>ao
2a' 2a '
Hence ^ = ±y/^ ± v^^^j .
Thus there are four real or imaginary roots.
Similarly, whenever an equation only contains the
unknown quantity in two terms one of which is the
square of the other, the equation can be reduced to two
alternative equations : for, whatever P may be,
aP"+6P + c =
is equivalent to P = — 5 ± ^^— ^ •
Zia Zxh
Ex. 1. To solve a?*  \^x^ + 9=0.
We have (a:«9)(a!«l)=0;
.•, ajSg^ giving «= ±3;
or else ««= 1, giving x= db 1.
Thus there are four roots, namely + 1,  1, + 3,  3,
Equations op moHEk degheb than the second. 135
Ex. 2. To soWe (aj2+a;)3+4 {x*+x)  12=0.
The equation may be written {x^+x + 6) (x^+x2)= 0.
H^oe a;«+a5+6=0, or 4?»+«2=0.
Therootsof a;2+a;+6=0 are 5±rV23.
Therootsof a^ + a;2=0 are 1 and 2.
Thus' the roots are 1, 2,  h **= on/ " 23»
Ex. 3. {sxi»+2)^ + 8x(x^+2) + 15a^=0.
The equation is equivalent to
(a;« + 2 + 5aj)(a5^ + 2 + 3a;) = 0.
The roots of ar» f 3aj+ 2=0 are  1 and  2.
^ 5 /i 7
Therootsof «» + 5a; + 2=0 are 2'*=^2^*
Thus the equation has the four roots  1,  2,  5 "*= n \/l7.
Ex.4. To solve ax'^+bx + c+p Jax^hbx+c + q=0.
Put y=Jax^\rhx\rC't
then y*+py+g=0,
whenoe we obtain two values of y, a and /3 suppose.
We then have aa^ + bx\cssa\
or aa;*+6aj + c=j8^,
and the four roots of the last two quadratic equations are the roots
required.
Ex. 6. To solve 2a?' 4a;+3iv/a;*2a?+6=16.
The equation may be written
2(ai«2aj+6) + 3V(a;'2aJ + 6)27=0.
Put y=ij(x^  2fl5+ 6) ; then we have 2y3+ 33^  27=0,
whence y=3, orys^.
Hence aB?2«+6=9, giving a8=3 or 1;
81 1 »—
orelse rr' 2a5+ 6= j, giving «ssl±^ /^/el.
Thus the roots are 8; 1; l^^J^,
136 EQUATIONS. ONE UNKNOWN QUANTITY.
9
Ex. 6. To solve (a: + a) (aj + 2a) (x + 3a) {x + 4a) = y^ a*.
Taking together the first and last of the factors on the left, and
also the second and third, the equation becomes of the form we are
now considering. We have
9
{pi^+5ax+4a^) (a;»+5ax+6a*)=Yg«*'
9
Hence (ar» + 5ax)^ + 10a* (x^ + Box) + 24a*= jg a\
,',x^+ 5ax = T a^, or else x^ + 5ax =ja\
4 4
■J »»
Hence a;+^a=0, or ir + ^a=db<^10.
Thus the roots are "" o ^> "*1^ ^ Q '^^^ •
V^136. Reciprocal Equations. A reciprocal equa
tion is one in which the coefficients are the same whether
read in order backwards or forwards ; or in which all the
coefficients when read in order backwards differ in sign
from the coefficients read in order forwards. Thus
ax* + 6a;* f ca?' + 6a? 4 a = 0,
and OjX^ \hx* \ cx^ — CO? — hx — a =^
are reciprocal equations.
Ex. 1. To solve
ax^ + 6a;* + 6a; + a=0.
We have
a(a;3+l) + 6a;(a; + l)=0,
that is
(a; + l){a (ar» x + 1) + 6a;} =0.
Hence
«=!,
or else
aa;* + (6a)a;+a=0.
Ex. 2. To solve
aa;* + 6a;3 + ca;* + 6a; + a = 0.
Divide by x^;
then we have
''('^"'i)"'^^''^)'"'""'
Now put
»+i=y;
then
x^+l=y*%.
Hence a(y«2) + 6y+c=0.
ROOTS FOUND BY INSPECTION. 137
Let the two roots of the quadratic in y be a and /3; then the
roots of the original equation will be the four roots of the two
equations
x+=a and x + =3.
X X
Ex. 3. To solve ax^ + bx* +ca^cx^bxa=0.
We have a(a:»l) + 6a:(x»l)+ca;*(a;l)=0,
that is (xl){a(x*+x^+3^+x + l) + bx(al*+x+l) + cx^}=0.
Hence a;=l, or else
aa:*+(6 + a)a» + (a + 6+c)a;*+(6 + a)a; + a=0.
The last equation is a reciprocal equation of the fourth degree
and is solved as in Ex. 2.
/137. Roots found by inspection. When one root
of an equation can be found by inspection, the degree of
the equation can be lowered by means of the theorem of
Art. 88.
Ex. 1. Solve the equation
x(a:l)(a;2) = a(al)(a2).
One root of the equation is clearly a. Hence a;  a is a factor of
x{xl)(x2) a(a  1) (a  2), and it will be found that
x{x'l){x2)a(al){a2) = {xa){x^{da)x+(al){a2)}.
Hence one root of the equation is a, and the others are given by
aj«(3a)x+(al)(a2) = 0.
Ex. 2. Solve the equation
a^{2x^''llx+Q=0,
Here we have to try to guess a root of the equation, and in order
to do this we take advantage of the following principle : —
If a;= ± 3 be a root of the equation ax^+ bx"~^ + ... + fc=0, where
P
a, &, ... ^ are integers and ^ is in its lowest terms, then a will be a
p
factor of k and /3 a factor of a. As a particular case, if th^e are any
rational roots of «**+... + A; =0, they will be of the form x=^a,
where a is a factor of k.
In the example before us the only possible rational roots are ± 1,
±2, dbS, and ^6. It will be found that x=2 satisfies the equation,
and we have
(a;2)(a;2 + 4a;3)=a:8 + 2aj>llaj+6.
Hence the otiier roots of the equation are given by
«« + 4a;3=0,
and are therefore  2 ± ^7,
138 EQUATIONS. ONE UNKNOWN QUANTITY.
Ex. 3. Solve
{a'x)*+{xb)*={ah)\
Sinoe x=a and x=b both satisfy the eqnation, {x  a) (a;  b) will
divide {oaj)* + (a:6)*(a6)*, and as the qnotient will be of the
second degree, the equation formed by equating it to zero can be
solved.
We may however proceed as follows. The equation may be written
{ax)*+{xb)*={{ax) + {xb)}^
= (aa:)* + 4(ax)3(a;.6) + 6(aa:)«(x6)a
+4(aaj){a;6)»+(aj6)*;
.. 2(aa;)(aj6){2(aa;)a + 8(aa;)(a56) + 2(a;6)«}=0.
Thus the required roots are a, b and the roots of the quadratic
a?a;(o + 6) + 2a»3a6 + 26«=0.
Ex. 4. Solve the equation
(g6)(gc) y (a?c)(a;a) A^a){X'b)
^ (a'^iacy"^ (bc)(bay^ {ca)(cb)~^'
The equation is clearly satisfied by d;=a, by fl;=&, and by a;=c.
Also, since the coefficient of nfi is zero, the sum of the roots is zero.
[Art. 129.] Hence the remaining root must be abc. '
Thus the roots are a, 6, c,  (a+b+c).
V 138. Binomial Equations. The general form of
a binomial equation is a?*" + & = 0.
The following are some of the cases of binomial
equations which can be solved by methods already given —
for the general case De Moivre's theorem in Trigonometry
must be employed.
Ex.1. To solve a:»l=0.
Since Q^''l=(x'l){pi?+x + l\
we have a? 1=0;
or else oe^ + a; + 1 = 0, the roots of which are
2*'"2~*
Hence there are three roots of the equation 0:^=1 ; that is there
are three cube roots of unil^, and these roots are
1 1 ro ^ 1 1 /— o
CUBE ROOTS OF UNITY. 139
Ex. 2. To solTe a:*  1 =0.
Since x*  1 = (a;  1) (aj + 1) (x + ^/^) (x  V^), the four fourth
roots of unity are
1, 1, /^Tl and ^1.
Ex. 3. To solve x»  1=0.
Hence xs=l;
or else a5*+x*+a;^+a; + l=0.
The latter equation is a reciprocal equation. Divide bj x^, and
we have
a»+::«+a;++l=0.
or X
Put a?+=y;
then a;* + 3=y*2;
.•. y22+y + l=0;
1^ Vs
•••y=2='^
Hence a; +  = n^>
a; 2
thatis a;3a;^^i^^ + l=0.
Hence g^"^ ^^^ =fcj 7102^5,
or ^^ lV5 ^^^_;^Q^2^/5,
or « = !.
Ex.4. To solve a:* +1=0.
a;*+l = (ar» + l)»2.r2=(x«+lV2x)(a;« + l+^2a;).
Hence a;^=FV2a; + l=0;
l/ 139. Cube roots of unity. In the preceding article
we found that the three cube roots of unity are
An imaginary cube root of unity is generally repre
sented by ©; or, when it is necessary to distinguish
» » X ~~ ■
140 EQUATIONS. ONE UNKNOWN QUANTITY.
between the two imaginary roots, one is called cd,, and the
other CD,, so that 1, ©j and o), are the three roots of the
equation aj^ — 1 = 0.
Taking the above values, we have
also co,(o, = i( 1 + J^) { 1 J^) = 1.
These relations follow at once from Art 129; for the
sum of the three roots of .i?' — 1 = is zero, and the
product is 1.
Again a,,' = i(l + Vr3)»=(l/l3) = a,„
and < = i(lN/^)' = Kl + N/^) = «P
so that (0^=0)^ and (o^=fo^. These relations follow at
once from
6)j(»j = 1 and a)j' = (o^ = 1.
Thus if we square either of the imaginary cube roots of
unity we obtain the other.
Hence if cd be either of the imaginary cube roots of
unity, the three roots are 1, cd and a}\
We know that
Hence a+b + ciB a faotorofa'+&'+c'3a&c, and this is the case
for aU values of a, &, c.
Hence a + («6) + i^c) is a factor of a« + («6)« + (w^c)'  3a («6) (w«c),
that is of a'+&^+c^3a5c; and a+icr^h + <ac can similarly be shewn
to be a factor.
Hence a'+6' + c'3a6c=:(a+6+c)(a+w6+«*c)(a+w''6+wc).
EXAMPLES XI.
Solve the following examples :
1. aj*2ic«8 = 0.
2. «• + 7aV  8a« = 0.
EXAMPLES.
3.
x^  7aV  8a« = 0.
4
X «* + 1 5
ic' + l ' a; 2*
^'
iB" + 2 iB" + 4aj+l 5
aj*+4aj+l ' a;" + 2 2*
141
9.
/
10.
/
11.
/
12.
13.
/
14.
15.
16.
17.
/
18.
/
^20.
22.
23.
24.
25.
26.
6. (iB' + a;+l)(a;» + a; + 2) = 12.
7. (ic« + 7a; + 6)'3iB»21aj=19.
8. ^/167a:aJ"=a;" + 7aJJ.
6 \/a;"  2aj + 6 = 21 + 2a;a5'. .
(a  1)(1 +a; + a;*)» = (a+ 1) (1 +aj* +aj*).
(a;+l)(aj + 2)(a: + 3)(a: + 4) = 24.
(aj + a) (a; + 3a) (a; + 5a) (aj + 7a) = 384a*.
(a:  3a) (a:  a) (aj + 2a) (a; + 4a) = 2376a\
(iB+ 2)(a; + 3) (aj + 8) {x+ 12) = 4a^.
2jk« 3a;  21 = 2a; Ja^  3a; + 4.
a;*  2 (a + 6) a;« + a* + 2a6 + 6" = 0.
a;*  2a; V  2a;^6» + a* + 6*  2a"6* = 0.
4a;*4a;'7a;*4a; + 4 = 0.
9a;*  2 4a;»  2a;"  2 4a; + 9 = 0.
a;^ + l=0. /2I. a;«l=0.
3a;»14a;« + 20a;8 = 0.
a;*  15a;" + 10a; + 24 = 0.
a;* + 7a;'7a;l = 0.
{xay {bcy + (xby (cay + {xcy (a6)' = 0.
a;(a;l)(a^2) = 9.8.7.
142 EQUATIONS. ONE UNKNOWN QUANTITY.
27. a;(ajl)(aj2)(aj3) = 9.8.7.6.
28. (a<+ (6a:)»= (a + 6 2aJ)^
29. (aa;)* + (6a;)* = (a + 62a;)\
30. {axY + (6a;)' = (a+ 6 2aj)*.
31. ^aa; + IJhx = l/a^b2x.
32. ^a  x + ^bx = t^a + 6  2a;.
33. (a  aj)' + (aj  6)* = (a  6)^
34. /^a — a; + Jx  6 = Ja — h,
36. J^a  a; + ^a;  6 = ^a  6.
36. a;* + (aa;)* = 6V
37. (a; + a)* + (a; + 5)* =17 (a 6)*.
38. ^a; + ^ax= */b.
39. ofta; {x^a + by  {ax + bx\ ahf = 0.
40. obex (a; + a + 6 + c)'  (a;6c + a;ca + xab + a5c)* = 0.
(axy + (x^by __ a* + b^
^' (a + 62a;)« " {a + by
42. a;* + 6(a + 6)a»+(a62)6V(a + 5)6"aj + 6* = 0.
43. {of + 6')' = 2aa;' + 2ah'x  aV.
44. (a; + 6 + c) (a; + c + a) (a; + a + 6) + a6c = 0.
45. X + + n — +3=0.
b+c—x c+a—x a+o—x
4B (a^^g)' , (a;5)' (a;c)« ^
^' {x^ay{bcy ■*■ (aj6)«(c.a)» ■*" (a;c)»(a6)« *
4.7 (a; + g) (a; f b) (a?  a) (a?  6) _ (x + c) (a; t o?)
(a;  a) (a;  b) (x + a) {x + b)~ {x c) {x  d)
^ {xc)(x'd )
{x + c) (a; + c?) '
/;
CHAPTER X
Simultaneous Equations.
140. A SINGLE equation which contains two or more
unknown quantities can be satisfied by an indefinite
number of values of the unknown quantities. For we can
give any values whatever to all but one of the unknown
quantities, and we shall then have an equation to deter
mine the remaining unknown quantity.
If there are two equations containing two unknown
quantities (or as many equations as there are unknown
quantities), each equation taken by itself can be satisfied
in an indefinite number of ways, but this is not the case
when both (or all) the equations are to be satisfied by the
same vahies of the unknown quantities.
Two or more equations which are to be satisfied by the
same values of the unknown quantities contained in them
are called a system of simultaneous equati(ms.
The degree of an equation which contains the unknown
quantities a?, y, z.,. is the degree of that term which is of
the highest dimensions in x^y.z
Thus the equations
ax + d^y + d^z = a*,
a^ + y* + 2^ — ^xyz = 0,
are of the first, second and third degrees respectively.
V
144 SIMULTANEOUS EQUATIONS OP THE FIRST DEGREE.
141. Equations of the First Degree. We proceed
to consider equations of the first degree, beginning with
those which contain only two unknown quantities, w and y.
Every equation of the first degree in x, y, z,... can by
transformation be reduced to the form
cuv + by •^cz + ...=k,
where a^h, c, .,.k are supposed to represent known quan
tities.
Note. When there are several equations of the same
type it is convenient and usual to employ the same letters
in all, but with marks of distinction for the diflferent
equations.
Thus we use a, 6, c... for one equation; a\ h\ c'... for
a second ; a", V\ c"... for a third ; and so on. Or we use
ttj, 6j, Cj for one equation ; a,, 6,, c^ for a second ; and so
on.
Hence two equations containing x and y are in their
most general forms
ax\hy=c,
and a'x f Vy = c\
and similarly in other cases.
142. Equations with two unknown quantities.
Suppose that we have the two equations
ax\hy = Cf
and dx + h'y = c^
Multiply both members of the first equation by h\ the
coefficient of y in the second ; and multiply both members
of the second equation by ft, the coefficient of y in the
first. We thus obtain the equivalent system
ah'x + hh'y = cb\
a'bx + hh'y = ch.
SlMULTANEOX^S EQUAtlONS OB* tHE I'IRST DEGREK 145
Hence, by subtraction, we have
ch' — cb
whence x
ab'a'b'
Substitute this value of x in the first of the given
equations; then
cb'  a ,
""W^Tb^^v'^
whence y =
oc' — a*c
ah' a'b •
The value of y may be found independently of x by
multiplying the first equation by a and the second by a \
we thus obtain the equivalent system
a'dx + a'by = a'c,
a' ax + al/y = oc',
Hence, by subtraction, we have
{a'b — ah') y = a!c'ac'\
•'• y'^bah''
which is equal to the value of y obtained by substitution.
Note. It is important to notice that when the value
either of a? or of y is obtained, the value of the other can
be written down.
For a and a' have the same relation to x that b and V
have to y; we may therefore change x into y provided
that we at the same time change a into 6, b into a, a into
6', and 6' into a\ Thus from
c6' — c'6 , ca' — c'a
X = ^n Tr we have v= 7—7 — ft .
s. A. 10
14G SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE.
It will be seen from the above that in order to
solve two simultaneous equations of the first degree, we
first deduce from the given equations a third equation
which contains only one of the unknown quantities ; and
the unknown quantity which is absent is said to have
been eliminated.
143. From the last article it will be seen that the
values of x and y which satisfy the equations
cw? + Jy = c,
and a'x+Vy^c',
can be expressed in the form
^ ^ y ^ i
he' — b'e ca' — c'a dl — alb
So also, from the equations
CM? + 6y + c = 0,
and a'rc + Vy + c' = 0,
'f •
we have
^ y _
he' "Ve ea—c'a aU—ab*
It is important that the student should be able to
quote these formulae.
Ex. 1. Solve the equations
3a; + 2y = 13,
and 7«+3y=27.
X y 1
We have
2. 278. 13 "13. 727. 8 "3. 37. 2»
thatis fsi!La=?:;
15 10 5*
• • X = v" = «J>
o
and y = — = 2,
SIMULTANEOUS EQUATIONS OF .THE FIRST DEGREE. 147
Ex. 2. Solve the equations
X y
and =  7.
X y
These may be considered as two simultaneous equations of the
first degree with  and  as unknown quantities.
X y
We therefore have
1 1
X . y 1
8(_7)(6)2""2.2(7)4~4(6)2.3'
1 1
a? y _£.
11 52""26'
that is
1_ 11 _ 26
•'•sc"26' "^^ ^"n*
1 32 18
^^^ y = 26'^'^=i6
Kz. 8. Solve the equations
xy=ah,
aaj&y=2(a«62).
We have
X y 1
2{a^b^) + b(ab) a(ab)2{a^ly^) 6 + a*
thatifl . ^  y  ^ .
.*. ar= r = & + 2a:
6a
262a6a»
and w = r = a+26.
Instead of referring to the general formulae of Art. 143, as we have
done in the above examples, the unknown quantities may be elimi
nated in torn, as in Art. 142 ; and this latter method is frequently the
simpler of the two. Thus in this last example we have at once, by
multiplying the first equation by a and then subtracting the second,
(6a)y=a(ad)2(a«62).
a2a5 + 26> _,
"^ ba
10—2
148 SIMULTANEOUS EQUATIONS OP THE PIKST DEGEEE.
Then x = {a\2b)+ab;
,\ x = 2a+b,
144. Discussion of solution of two simultaneous
equations of the first degree. We have seen that the
values of a and y which satisfy the equations
axiby—c (i),
and a'x + b'y=^c (ii),
are given by
(aV  a'b) x^cb' — cb (iii),
(ba' 'b'a)y = ca' — ca (iv).
Thus there is a single finite value of x, and a single
finite value of y, provided that ab' — ab + 0.
If ah' — a'b = 0, a? will be infinite [see Art. 118] unless
cb' — c'6 = ; and, if ah' — a'b and cV — &b are both zero,
any value of x will satisfy equation (iii).
So also, y will be infinite if ah^ — a'b = 0, unless ca^—c'a
is also zero, in which case any value of y will satisfy
equation (iv).
If ah' — a'6 = 0, then ? =» t/ ; and if a5' — a'6 = and
a
also cb' — c'b = 0, then 7 = 77 =  .
a c
When equations cannot be satisfied hj finite values of
the unknown quantities, they are often said to be incon
sistent Thus the equations ax + by=^c and a'x + b'y = c'
are inconsistent if — , = 77 , unless each fraction is equal to
a b ^
7 , in which case the equations are indeterminajte. In fact
c
when 7 = T7 = — , it is clear that by multiplying the terms
of equation (i) by — we shall obtain equation (ii), so that
a
the two given equations are equivalent to one only.
SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE. 149
We have hitherto supposed that a, a\ 6, b' were none
of them zero. It will not be necessary to discuss every
possible case : consider, for example, the case in which a
and a' are both zero.
When a and a are both zero, we have from (i) y ^t >
6
and from (ii) yjf> These results are inconsistent with
one another unless r = n •
c c'
Hence, if a= a' = 0, and t—t,, the equations (i) and
(ii) are satisfied by making y — r, and by giving to x any
finite value whatever.
If however r=t=r>, the equations by = c and Vy = c'
cannot both be satisfied, unless they are looked upon as the
limiting forms of the equations ax+hy^c and a'x^Vy^c^
in which a and a' are indefinitely small and ultimately
zero. But from (iii) we see that when a and a' diminish
without limit, x must increase without limit, cV — c'b not
being zero. Thus, in the equations (i) and (ii), when a
and a' diminish without limit, and cV + c'6, the value of x
must be infinite. v /
Equations with three unknown quantities.
145. To solve the three equations :
(iX'{by\cz=^d (i),
a'x\Vy\c'z = df .(ii),
a''x^V'y + c"z = d' ,!(iii).
Method of successiye elimination. Multiply the
first equation by c', and the second by c ; then we have
m'x 4 bey + cc^z = dc\
150 SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE.
and a' ex + h'cy + c'cz = dc ;
therefore, by subtraction,
{ac — a'c) X + {be' — Vc) y = dc' — dfc (iv).
Again, by multiplying the first equation by c" and
the third by c and subtracting, we have
{ao"  a''c) X + (6c"  Vc) y = dc''  d"c (v).
We now have the two equations (iv) and (v) from
which to determine the unknown quantities x and y.
Using the general formulae of Art. 143, we have
_ ■, Q^c'  Vc) (dc'  d''c) + (dc  d'c) (6c"  Vc)
^ {ac'  ac) {be'  b"c)  {be'  b'c) {ae"  a"c) '
Method of undetermined multipliers. Multiply
the equations (i) and (ii) by X and jjl, and add to (iii);
then we have the equation
x{\a + fia +a") + y (Kb + lib' + b") + z (Xc + fie' \ c")
= (Xd + fid' + d"\
which is true for all values of X and fi.
Now let X and /^ be so chosen that the coefficients
of y and z may both be zero,
^, Xd + fid' + d"
then X = ^ — —^—f n
Xa\ fia + a
where X and fi are found from
Xb\fib' + b"^0,
and Xc + A6c' 4 c" = ;
. ^ _ M ^ 1
' ' b'c"  b"c " 6''c  be" " be'  b'c '
Hence
_ d {b'c" ~ 6V) + d' {b"e ~ be") + d" {be' b'c )
^ "■ a {Vc"  b"o') + a! {b"c  be") + a" (6c'  6'c) '
[The numerator and the denominator of the first value
of X, which was obtained by eliminating z and y in succes
." >
SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE. 161
sion, can both be divided by c ; and the two values of x will
then be seen to agree.]
Having found the value of x by either of the above
methods, the values of y and z can be written down.
For the value of y will be obtained from that of x by
interchanging a and b, a! and b\ and a" and 6". The
value of y can also be obtained from that of ^ by a
cyclical change [see Art. 93] of the letters a,b,c\ a\ b\ d ;
and a'\ V\ c"; and a second cyclical change will give the
value of z.
It should be remarked that the denominators of the
values of x, y and z are the same, and that there is a
single finite value of each of the unknown quantities
unless this denominator is zero.
Ex. 1. Solve the eqoations :
a; + 2y + 32; = 6 (i),
2a; + ^+ « = 7 (ii),
3a; + 2y+9« = U (iii).
Multiply (ii) by 3, and subtract (i) ; then
5x+10y=15 (iv).
Again multiply (i) by 3, and subtract (iii) ; then
4y = 4 (v).
From (v) we have y = 1 ; then, knowing y, we have from (iv) a? = 1 ;
and, knowing x and y, we have from (i) x; =1.
Thus xsysjSsl.
Ex. 2. Solve the equations :
a? + y+is=l (i),
ax + by + cz = d (ii),
a^x + bhf + c^z=cl^ (iii).
Multiply (i) by c and subtract (ii); then
{ca)x+(eh)y=cd (iv).
Again multiply (i) by c^ and subtract (iii) ; then
(caaa)a?+(c«63)yc«(i« (v).
Now multiply (iv) by c + 6 and subtract (v) ;
then
(c  a) (6 a)aj=(c  <J) (6  <i) ;
(6  o) (c  a) *
152 SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE.
The values of y and ;; may now be written down : they are
(C'd)(ad) _ (ad)(hd)
y (c^h)(a~hy ^~'(ac)(hc)'
Instead of going through the process of elimination, we may at
once quote the general formulae. Thus
{hc){hc + d{h + c)'d?}
(h'C){hc + a{h\'C)a^\
TL — w {* *s above.
(6  a) (c  a)
Ex. 3. Solve the equations :
x + y + iS=a + 6 + c (i),
ax\hy+cz = hc\ca + db (ii),
'bcx\cay + <ibz=%ahc (iii).
We have
(a + 6 + c) (aft*  oc*) + (6c + ca + a5) (ca  oft) + 3a6c (c  6)
al^ ac^ + a{ca'ah) + hc(ch)
__a(6c){(6 + c) (a\h + c) he  ca  ah %hc}
"" (6c) {ah + oc  a*  6c}
o(6c)«
"~ " (a  6) (a  c) *
The values of y and z can now be written down : they are
6(ca)« _ c (g  6)^
^" (6c)(6a)' ''"~(ca)(c6)'
Ex. 4. Solve the equations :
a;+ay + a^« + a'=0 / (i),
a; + 6y+6*x;+68=0 (ii),
a; + cy+c*«+c*=0 (iii).
The equations may be solved as in the preceding examples, or as
follows.
It is dear that a, 6, c are the three roots of the following cubic
in X
X'+2X« + yX+a;=0.
Hence from Art. 129, we have at once
iS=(a + 6 + c),
l/ = 6c + ca+a6,
f^n4 ^=a6c,
_ J
SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE. 153
146. Equations with more than three unknown
quantities. We shall return to the consideration of
simultaneous equations of the first degree in the Chapter
on Determinants, and shall then shew how the solution of
any number of such equations can be at once written
down.
The method of successive elimination or the method
of undetermined multipliers can however be extended to
the case when there are more than three unknown quan
tities. For example, to solve the equations
dx +by +CZ \dw =e (i),
a'x +b'y +&Z +d'w =e' (ii),
a''x +6'V +c'z +d"w ^^' (iii),
al^'x + V"y + c'"z + dl'^w = e'" (iv).
Multiply (i) by X, (ii) by /a, (iii) by i/, and add the
products to (iv). Then we have
X (aX + a> + a'V + a'") + y{b\ + Vfi + h"v 4 6'")
4 z (cX + c> + c'V + O + ii; (d\ + dffi + d!'v 4 d!")
= 6X + e>4e'V + e'" (v).
Now choose X, /a, i/ so as to make the coefficients of y,
z and w in the last equation zero ; then
e\+eii\ev + e ..v
X=— f ;; 777 (Vl),
where X, /a, v are to be found from the equations
6X + VfJL + b"p + y" = 0^
cX + c> +c''i/ 4c'" = ol (vii).
dX + d> + d'V + d''' = 0)
Hence we have to solve (vii) by Art. 145 and then
substitute the values of X, /a, and v in (vi) ; this will give
the value of x\ and the values of the other unknown
quantities can then be found by cyclical changes of the
letters, a, 6, c, d, &c/
154 SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE.
EXAMPLES XIL
Solve the following equations.
^ 3^6 = 2' ^' xy"^'
X ^y _\ ^^0.^10
5"10~2 X ^y"^^'
3. a; + =^, 4. +5 = + = 10.
y 2' X y X y 
y 3
5. ax + by = 2ab, 6. aj + ay + a* = 0,
hxa^ = b'(ir. x + by + b' = 0.
7. a; + y=2a, 8. (b + c)x + {bc)y = 2ah,
(ab)x=(a + b)y, {c + a)x + (ca)y = 2ac.
9. bx\ay = 2ah,
a*x + 5*y = a' + 6'.
10. (a + 6) a; + 6y = aaj + (5 + a) y = a* — 6^
11. x + y + z=ly 12. a; + y + 2; = l,
2a:+3y + ^ = 4, ? .^ . 4.^1
4a; + 9t/ + »=16. 2^4^*^" '
5 3 ;5 
13. x\2y + 3z = 3x + y+2z = 2x + Sy + z = 6.
y 14. y + 2 = 2a, . / 15. y + «  a; = 2a,
« + a; = 26, « + a;  y = 26,
aj + y=2c. a: + y» = 2c.
\
\
EXAMPLES. 155
16. y + z3x=2ay y\l. ax + by ¥cz=^ly
z + X'3y = 2b, bx + cy + az = l,
a; + y  3« = 2c. cx + ay + bz=l.
TO y + zx z^+xy _x + yz _
+ c c + a a +
^19. x + y + z=0, 20. x^y + z=^a + b + c,
ciX + by\c^=ly bx + cy + az = bc +ca+aby
a'x + b'y{'C'z = a + b + c. cx + ay+bz = bc + ca+al>,
21 x{y + z = a + b + c,
bx + cy + az = a^ + b^ + c\
cx + ay + bz = a^ + b' + c^,
22. x + y + z = Oy
(b + c)x + {c + a)y^{a + b)z = {bG){ca){a 5),
bcx + cay + ahz = 0.
V^23. a:s: + by + cz = a, 24. xay + a'za^ = 0,
bx+cy^az = by xby +b'z b"" = 0,
cx^ay + bz = c. x cy + c^'z c^=0.
25. ax + by^cz = 7n,
a'x + b'y + c'z = m*,
a^x + 6*y + c^z = m\
26. ax + cy + bz = a' + 2bc,
cx + by{az = b' + 2ca,
bx + ay + cz = c^ + 2ab,
27. x + y + z = 2a + 2b +2c,
ax + by + cz = 2bc + 2ca + 2a&,
(6 c) 35 + (c  a) y + (a 6) 2J == 0.
aa5 + 6y + c5J = a + 6 + c,
6caj + cay + ctbz = 0.
156 SIMULTANEOUS EQUATIONS OF THE SECOND DEGREE.
29. x + y + z^lhm\n,
1x ¥ my + 7m; = mn + 7i^ + hn^
{m—n)x+{nl)y\(lm)z = 0.
30. lx^ny + 7nz = nx + my + lz = mx ^hj{nz
= r + m* + w*  3 Imn,
31. Px + m'y + n*a = hnx + mny + nlz = wZa; + Imy + Tymjs
= ^ + m + w.
32. +^+ =1.
a + a a + p a + y
6 + a ft + jS 5 + y
* • y • " =1.
33. y + z + w=^a,
z + w + x = b,
w + x^y = c,
x + y + z = d,
34. x + ay + c?z + d^w + a* = 0,
aj + 6y + 5*2? + 6'i/? + 6* = 0,
x^cy ^ ^z ^ c"w + c* = 0,
x+dy + c^z\ <Pw + d^ = 0.
I
y Simultaneous Equations of the Second Degree.
147. We now proceed to consider simultaneous equa
tions, one at least of which is of the second or of higher
degree.
We first take the case of two equations containing two
unknown quantities, one of the equations being of the first
degree and the other of the second.
^J
SIMULTANEOUS EQUATIONS OF THE SECOND DEGBEE. 157
For example, to solve the eqoations :
8a?+2y = 7,
8a;«2y«=25.
From the first equation we have
^" 8 •
Snbstitate this yalne of a; in the second equation ; we then have
whence y* + 14y + 13 = 0,
that is (y + 13)(y + l)=0;
.'. 2/= 1, or y= 18.
Ify=1, «=^ = 3;
and if y= 13, a5=ll.
Thus ar=3, y=l; or a? =11, y=13.
From the above example it will be seen that to solve
two equations of which one is of the first degree, and the
other of the second degree, we proceed as follows : —
From the equation of the first degree find the value of
one of the unknown quantities in terms of the other un
known quantity and the known quantities; and substitute
this value in the equation of the second degree ; one of
the unknown quantities is thus eliminated, and a quadratic
equation is obtained the roots of which are the values of
the unknown quantity which is retained.
The most general forms of two equations such as we
are now considering are
Ix 4 my + n = 0,
aa? 4 hxy {• cy^ + dx + ey +/= 0.
From the first equation we have
my+n
a? ^ •
158 SIMULTANEOUS EQUATIONS OF THE SECOND DEGREE.
Hence on substitution in the second equation we have
to determine y from the quadratic equation
a {my 4 w)' — Iby (my + n) + cPy*
 dl {my + n) 4 ePy +fV = 0.
Having found the two values of y, the corresponding
values of x are found by substitution in the first equation.
148. It should be remarked that we cannot solve any
two equations which are both of the second degree; for
the elimination of one of the unknown quantities will in
general lead to an equation of the fourth degree, from
which the remaining unknown quantity would have to be.
found ; and we cannot solve an equation of higher degree
than the second, except in very special cases.
For example, to solve the equations
aa^ + bx + c = y, a?' + 2/* =d.
Substitute cw?* + bx + c toiym the second equation, and
we have
ai" + (aa? '{•bx + cy =^ d,
which is an equation of the fourth degree which cannot be
solved by any methods given in the previous chapter.
149. There is one important class of equations with
two unknown quantities which can always be solved,
namely, equations in which all the terms which contain
the unknown quantities are of the second degree. The
most general forms of two such equations are
aa? + bxy + cy^=^d
and a V + Vosy + c'y^ = d\
Multiply the first equation by d\ and the second by d
and subtract ; we then have
(ad'  a d) a;* + (bd'  6'd) xy + {cd'  c'd) / = 0.
SIMULTANEOUS EQUATIONS OF THE SECOND DEGREE. 159
The factors of the above equation can be found either
by inspection, or as in Art. 81 ; we therefore have two
equations of the form Ix + my = either of which com
bined with the first of the given equations will give, as in
Art. 147, two pairs of values of x and y.
Ex. 5. To solve the equations :
y*xy =15 (i),
x^+xy=14: (ii).
We have 14 [y^  ocy) = 15 (or* + xy) ;
.. 15x^ + 29xyUy^ = 0,
that is
^5x2y)(3a:+7y) =
Hence
5x2y:
=0,
or else
dx+7y:
= 0.
U 5x2y =
0,
we
have from (i)
y^ly^=
:15,
whence
2/= ±5.
Hence also
a;=±2.
If Sx + 7y =
.0,
we
have from (i)
y'+ly'=
15,
whence
y=
*i
and then
x=
7
7 3
Thus a:=±2, y==fc5; or ar=±^» 2^""*"^*
150. The following examples will shew how to deal
with some other cases of simultaneous equations with two
unknown quantities ; but no general rules can be given.
Ex.1. To solve a:y=2,
xy = 15.
Square the members of the first equation, and add four times the
second; then
160 SIMULTANEOUS EQUATIONS OF THE SECOND DEGREE.
Hence a; + y=±8,
which with a;y=2,
gives x=5 or 3,
and ^=3 or 5.
Thus «=5, y=3; orx=3, y=5.
Ex.2. To solve x^+£y+y^==a^7 (i),
«*+a:V+y*=6* (ii).
Divide the members of the second equation by the corresponding
members of the first; then
x^xy + y^=^ (iii).
From (i) and (iii) by subtraction we have
b*
2xy=a^^ (iv).
From (i) and (iv)
. „ , Sa*b*
. /3a*6* , ,
•••^ + y==^V~2^ ^''^*
From (iii) and (iv) we have
^ /36*a* . ..
•'•^"y==^\/^^i (^>
Finally, from (v) and (vi) we have
If / 3a* 6^ / 3Ma* 
"2] V 2a« V 2a« )•
1 r /3a* 6* /36*a*)
X
•
Ex. 3. To solve x^2y^=4y,
3aj»+a;y2y*=16y.
Multiply the first equation by 4, and subtract the second; then
that is (« + 2y)(a;3y)=0;
/. a; + 2y = 0,
or else aj8y=0.
Simultaneous eqi^atIonS of tHE second deg&ee. l6l
If x+^=sO, the first equation gives
/. y=0 or t/ = 2,
whence x=0 or x= 4.
If xdy=Oi the first equation gives
V2y«=4y;
.. y=:0 or y=^r
whence a;=0 or a?=5=.
Thus x=0, y=0; a;=:4, / = 2;
12 4
or a?=y, y=^.
Ex. 4. To solve x^+y^=(x + yhl)\
By subtraction we have
{x+y + lfixy +2)^=0,
that is (2x + 3)(2yl)=0.
Hence 2a5+3=0, or 2yl=0.
If 2x+3=0, we have
whence y =  2.
If 2yl=0, we have
,4
(■*■')'.
07=3
2
~3'
k
3
y=
:2;
2
y=
1
"2
whence
Thus
or
Ex.6. To solve »+y=26,
a^+y*=2a*.
Put 85=6+2; then, from the first equation, y=6j.
S. A. 11
162 SlMULTAlfEOliS EQUATIONS OP* THE SECOND DEG&Efi.
Hence (6 + «)* + (6  «)* = 2a\
irhenoe after reduction
Thus a;=6dL^{36»i78&*+a*},
and y=6=P/s/{S6a± */§&*+«*}.
EXAMPLES XIIL
Solve the following equations ; —
1. iB + y = jB»/=23.
2. iB*43^' + ic + 3y = 2iBy=l.
3.
x* + xy=12f
xy2y'=l.
4.
3y"a^iB'=17.
\
5.
xy = 5,
1 1 5
y a? ~ 84 •
6.
a5 + y = a + 6,
a b ,
V + =1.
0? + 6 y +a
7.
a(« + y) = ^(ay) =
= xy.
a
1 1 1
a* xy~ a'*
1 1 1
y* yas 6* '
9.
xy
10.
x + y=2a,
iB» + y» = 25».
11.
iB"ajy + y*=109,
a?* + iB'y' + y*=4251.
12.
af'hxy + 2^ =
133,
13.
a; + y=72,
aj + Vay + y =
19.
^aj+4^y = 6.
J
EXAMPLES. 163
14. i+=2, 15. x^y=l,
16. a^ + y^ + ^xy 4(a5 + y) + 3 = 0,
a^ + 2 (cc + y)  5 = 0.
17. ;K? + xy + x=\4:, 18. af + y^=^9y
y* + ay + y = 28. a^^xy^y^ = 3.
19. a; (y — 6) = y (a;  a) = 2a5.
20. a; +  = l, 21. ax^by=2ab,
y
1 .
y+ = 4.
22. ^ + 2^=12,
y X
1 1 1
a? y 3
Of.
24. Qcy = ay
y
ajy^ = .
^ a; a
26. a; + y = 6, 27, aj + y=8a;2^,
(a» + y*) («» + y') = 1440. «;• + y" « 40aj'y«.
28. aj"ajy = 8aj+3, 29. ^iil = 3,
a^y« = 8y~6. gy _1
l + osy 3 *
11?
 4
y
5_
a;
2.
23.
x^
— 4
y
■a^
= a»,
X
xy
= 5".
25.
aj +
y +
a;
14.
aj'+y*
= 84,
164 SIMULTANEOUS EQUATIONS OB* THfi SECOND DfiGHEfi.
80. a,y = «(a^A 31. M=^f,
32. ?^«=y+^=?+?.
a X b y y X
151. Equations with more than two unknown
quantities. No general rules can be given for the solu
tion of simultaneous equations of the second degree with
more than two unknown quantities : all that can be done
is to solve some typical examples.
Ex. 1. Solve the equations :
(x+y)(x\z)=sa* (i),
{y+z)(y+x)=^}fi (u),
(z+x)(z\y)=sc^ (iii).
Multiply (ii) and (iii) and divide by (i) ;
then (y+«)"=;5;
Similarly we have
a'
/. y+«=± — (iv).
_ ah ...
and a;+y=± — (vi).
Also from the original equations it is clear that the signs must aU
be positive or all be negative.
Add (v) and (vi) and subtract (iv) from the sum; then
. fea ah bc\
\b c aj
2ahc
• • flj^s sfc
BoalBo y=* ^^ ,
and «=± 2^j3
SIMULTANEOUS EQUATIONS OF THE SECOND DEGREE. 165
Ex. 2. Solve the equations :
x(y+z)=a (i),
y{z+x)=b (u),
z(x{y)=c (iii).
We have y{z + x){z(x+y)x{y+z)=b{ca,
that is 2yjr=d+ca.
Similarly 2zx=se+ah,
and 2xy^a+b e.
Hence P^) (^^^) « («l&c)(c+a^ .
2yz b+ca *
•• ^^ (b^ca)
Hence ^=^ /fe±Z«i^±^),
and similarly t/ = ± / ^^ — ^^^ — — — = .
»ixu omixxiwrijr y *^ 2(c + a6) '
and ,= =, /(lliZ^^lizl).
\/ 2(a + 6c)
Ex. 3. Solve the equations :
x2+2y«=a (i),
y^ + 2zx=a (ii),
z^+2xy = b (iU).
By addition {x+y + z)*=2a + b;
/. x+y+z=^ :kj2a + b (iv).
From (i) and (ii) by subtraction
{x'y){x + y'2z) = 0.
Hence x=y (v),
or else 4?+y2i?=0 (vi).
I. If x=y, we have £rom (ii) and (Hi) by subtraction
z^+a^2xz=ba;
/. zx^^fjba (vii).
Hence, from (iv), (\) and (vii),
166 SIMULTANEOUS EQUATIONS OP THE SECOND DEGREE,
n. When x+y2z—0, we have from (iv)
2 I
and x + y=^^J2a+b.
Also, from (ii), y * + »(a; + y) = a,
which with the preyious equation gives
and y = ▼ k/^Y~ * 8 >/^*+**
Ex. 4. Solve the equations :
We have l^z + ehf=xyz (i),
€^x + a^z=xyz (ii),
and ahf\l^x=xyz (iii).
Multiply (i) by  a*, (ii) by 6>, and (iii) by c*, and add ;
then 2b^ehi={a'+b*+e^)xyz.
Henoe x=0,
26»c«
or else yz^ — ^ . i>< . ^ •
If x=0, y and z must also be zero.
Henoe a:=5f=«=0;
26«c«
and similarly *^^ c»+a«d^ •
2a26a
The solution then prooeeds as in Ex. 2,
Ex. 6. Solve the equations :
a?yz^ckf
y^^zxssb,
z^xys=e.
BIMULTANEOUS EQUATIONS OP THE SECOKD DEGREE. 167
We have (x^^yxf'{y^zx){»*'iKy)=a^'be,
that is X («* +!(• +fi^  ^xyz)^a^  he.
Hence, from the last equation and the two similar onea^
g _ y __ g
a*  6c l^ca c^ab'
Hence each fraction is equal to
/ ^y* I 1
Ex. 6. Solve the equations :
aj+y+2=a + 6+c (i),
ar^+yHif*=a«+6« + c2 (ii),
5 + F + c=« (•")•
It is obvious that a;=a, y=6, 2=c will satisfy the equations : put
then a;=a + X, y=&+fi, iE=c+v, and we have after reduction
X+ft + v = (iv),
 + ^ + =0 (v),
2(a\+6/A + CJ')+X2 + /A«+y«=0 (vi)
From (iv) and (v)
X _ /t __ 9
a(6  c) "" 6(c  o) "" c (a  6) *
whence from (vi)
X _ 2(5c)(ca)(ary>)
Hence d;=ay y=5, 2s=e;
or else
gg __ y6 __ «(? 2 (6  c) (c  a) (a  6)
a(6c)'"6(ca)~c(a6)'"a*(6c)2 + 62(ca)2 + c«(a6)'S*
Ex. 7. Solve the equations:
x+y+z= 6,
a:yzs 6.
This is an example of a system of three symmetrical equations.
Such equations can generally be easily solved by making use of the
relations of Art. 129. Thus in the present instance it is clear that
jc, y, c are the three roots of the cubic equation
\»6XHllXQ=0,
168 SIMULTANEOUS EQUATIONS OF THE SECOND DEGREE.
The roots of the cubic are 1, 2, 3.
Hence a:=l, y=2, z=B; or 4J=1, y = 3, «=2; or a;=2,
y=3, z=l; &C.
Ex. 8. Solve the equations :
x + y + z = a (i),
 +  +  =  (u),
X y z a ^ '*
yz+zx+xy=^ c^ (iii).
This again is a system of symmetrical equations, and two of the
relations of Art. 129 are already given ; we have therefore only to find
the third.
We have from (ii),
yz + zx + xy ^ 1^
xyz "" a *
.'. xyz= cu^ (iv).
Then, from (i), (iii) and (iv), we see that x^ y, z are the roots of
the cubic X^  aX«  c^\ + oc* = 0,
that is X«(Xa)c2(\a)=0;
.*. X=a, or X=±c.
Thus a?=a, y=c, z=^e; <fco.
Ex. 9. Solve the equations :
x^yz)=:a^bc), f
y^{zx)=h'^(ca),
z^{X'y)=(^{ah),
By addition
x^{yz)+%/*(Z'x)+z\xy)=a^(hc)\h^{ca) + c^{ah),
that is (y  «) (^  x) (xy) = {p c) (c a)(a 6).
By multiplication
i^H^(y~z)(zx){xy)=:a^h^c^{hc)(ca)(ah)\
:. x^H^^a^lf^cK
Hence xyz=sabc (i),
or xyz= abc (ii).
Again a*(bc)y + b^{ca)x=:aPy{y^z) + xy^(zx)
=xyz(yx) (iii).
Hence, if xyz=abc, we have from (iii)
{b*(ca){dbc}x{{a*{bc)abc}y = 0,
that is bx {be + ca' ab)  ay {be + ca  a6) = ;
x y z
t*.  = ? I and therefore each =  •
a Q
EXAMPLES. 169
Thus, when xyz^dbc. we have  = f =  ,
^ a c
Hence each is equal to . / y = ^1.
Thus = = =1. or *=f = i=l.
a o c ata o<a cuf
X _y f__
If xyz =  dbc, we have from (iii)
X V
 {he  ca'^db)=:j {ca  a6  6c).
Hence also each = (oJ  ftc  ca)
=<y {  (6c  ca  a6) (ca  o6  6c) (o6  6c  ca)}.
EXAMPLES XIV.
Solve the following equations :
1. yz = a\ 2. x(x + y + z) = a*y
xyc^, z{x\y + z)^ c\
3. ''yz + zx + xy = a, 4. yz = a(yhz),
yz — zx + xy = bf zx = b{zk'X),
yz + zx — xy = c. xy=^c{x + y).
6. yz = by + cz, 6. x'+2yz = 12,
zx = cz+ax, y'+2zx = l2y
a:y = ax + by. s^ + 2ocy=l2,
7. {y + z){x + y + z) = a, 8. {y + b){z + c) = a%
{z + x){xhy + z) = b, {z + c) (a? + a) = b',
{x^y){x + y + z) = c. (x + a) {y + b) = c^.
170 SIMULTANEOUS EQUATIONS OF THE SECOND DEGREE.
9. a^(yzy = a', 10. x(y + zx) = a,
y'{z'xy = b\ y(z+X'y) = h,
z'{xyy=(f, z^x + yzj^ c.
y+z z+x x+y
11. = —7— = = 2osyz,
12.
y f g z + x _x + y _a^ ^i^ + z*
a ~" b " c ~ a* + 6* + c* *
16.
I 3
a' + y ' = 2.
. 7
« + »"* = 4.
13. y« = a + y + «, 14. yz = a(y + z) + ay
zx = b + z¥x, zx=:a(z + x) + p,
xy^c + x + y. xy = a(x + y) + y.
15. yzf'^^cy^bz,
zxff'^az + cx,
xy — h* = bx^ ay.
17. « + y + 2?= 6, 18. aj + y + »= 15,
ixf + y' + z'=^U, iB» + y» + »»=495,
33^2 = 6. xyz = 106.
19. xhy + z^ 9, 20. aj+y + «= 10,
fic* + y* + «*= 41, y» + aB + a5y= 33,
(x? + ^'{z*^ 189. (3/+»)(2+a;)(ic+y) = 294.
21 yg _ gsc _ gy ag* + y* + g *
fe» + cy CSC + a2 ay + bx a' + b' + c*'
22. +?+?=!, 23. ««=?^ + ?,
03 6 c z y
x b z . ^ z X
+ + =1, 6y = +,
aye ^ X z
— +7+=l. C8f = H — .
a b z y X
EXAMPLES. 171
24. y* + »*a;(y + 2;)t=a*,
35* + y* — « (a5 + y) = c'.
25. aj* + y«  a* = y* + «aj  5* ^^*f a^y  c' = ^ (a* + y' + «*).
26. aj (a; + y 4 »)  (y" + «• + y») = a,
y (aj + y + ») — («* + a:" + ««) = b,
« (a? + y + »)  (a;* + y* + a;y) = c.
27. aj4y + « = a + 6 + c,
»* + y* + a* = o" + 5' + c%
(6 — c) aj + (c  a) y + (a  6) 2; = 0.
28. (aj + y) (a; + ») = aa:, 29. afyz=^ax,
(y+«)(y + «)=^y> y"«aj = 6y,
(» + aj) (« + y) = c«. s^"xy = cz.
30. a;' + a(2a; + y4«) = y' + 6(2y + « + aj)=«' + c(2£r + aj + y)
= (a; + y + «)•.
31. ^■^yz + s^^a'y
j5* + «aj + a;' = 6*,
05^ + ajy + y* = c*.
32. a'a; + 6"y + c'« = 0,
(6c)'^(ca)'^(a6)«^Q
ax by cz ^
111111
— + — + — =+ + .
yz zx xy a c
CHAPTER XI.
Problems.
152. We shall in the present chapter consider a class
of questions called problems. In a problem the magni
tudes of certain quantities, some of which are known and
others unknown, are connected by given relations; and the
values of the unknown quantities have to be found by
means of these relations.
In order to solve a problem, the relations between the
magnitudes of the known and unknown quantities must
be expressed by means of algebraical symbols : we thus
obtain equations the solution of which gives the required
values of the unknown quantities.
It often happens that by solving the equations
which are the algebraical statements of the relations
between the magnitudes of the known and unknown
quantities, we obtain results which do not all satisfy the
conditions of the problem. The reason of this is that in a
problem there may be restrictions, expressed or implied,
on the numbers concerned, which restrictions cannot be
retained in the equations. For example, in a problem
which refers to a number of men, it is clear that this
number must be integral, but this condition cannot be
expressed in the equations.
Thus there are three steps in the solution of a problem.
We first find the equations which are the algebraical
expressions of the relations between the magnitudes of the
1>110BLEMS. 173
known and unknown quantities; we then find the values
of the unknown quantities which satisfy these equations ;
and finally we examine whether any or all of the values
we have found violate any conditions which are expressed
or implied in the problem, but which are not contained in
the equations. The necessity of this final examination
will be seen from some of the following examples of
problems.
Ex. 1. A has £5 and B has ten shillings. How much must A give to
B in order that he may have just four times as much as J3 ?
Let X be the mmber of shillings that A gives to B,
Then A "wiU have 100  x shillings, and B will have 10 + a: shillings.
But, by the question, A now has four times as much as B.
Hence we have the equation
100a;=4(10+a;);
.•. a; =12.
Thus A must give 12 shillings to B.
It should be remembered that x must always stand for a number.
It is also of importance to notice that all concrete quantities of the
same kind must be expressed in terms of the same unit.
Ex. 2. One man and two boys can do in 12 days a piece of work
which would be done in 6 days by 3 men and 1 boy. How long
would it take one man to do it?
Let a;=the number of days in which one man would do the whole,
and let ^ssthe number of days in which one boy would do the whole.
Then a man does  th of the whole in a day ; and a boy does  th
X y
of the whole in a day.
By the question one man and two boys do ^th of the whole in a
day.
Hence we have
1 2_J^
«'*"y""12*
We have also, since 3 men and 1 boy do th of the whole in a
day,
X y ^
Whence a; =20.
Thus one man would do the whole work in 20 days.
1 74 PROBLEMS.
Ex. 3. In a certain family eleven times the number of the children is
greater by 12 than twice the square of the .number. How many
children are there?
Let z be the number of children ; then we have the equation
llx=2a?+12,
or 2a:'llaj+12=0,
that is (2ar3)(x4)=0.
Hence x=4, or a?=f .
The value 2t;=f satisfies the equation^ but it must be rejected, since
it does not satisfy all the conditions of the problem, for the number
of children must be a whole number.
Thus there are 4 children.
Ex. 4. Eleven times the number of yards in the length of a rod is
greater by 12 than twice the square of the number. How long is the
rod?
This leads to the same equation as Ex. 8; but in this case we
cannot reject the fractional result. Thus the length of the rod may
be 4 yards, or it may be a yard and a half.
Ex. 5. A number of two digits is equal to three times the product of
the digits, and the digit in the ten's place is less by 2 than the digit
in the unit's place. Find the number.
Let X be the digit in the ten's place ; then x+2 will be the digit
in the unit's place. The number is therefore equal to
10x+(a?+2).
Hence, by the question,
10a;+(x + 2)=:3x(a? + 2);
/. 3a» 6a? 2=0,
or (a?2)(3«+l)=0.
Hence «=2, orx=i.
Now the digits of a number must be positive integers not greater
than nine; hence the value x= •! must be rejected. The £git in
the ten's place must therefore be 2, and the digit in the unit's place
must be 4. Hence 24 is the required number.
Ex. 6. A number of two digits is equal to three times the sum of the
digits. Find the number.
Let X be the digit in the ten's place, and y the digit in the unit's
place; then the number wdl be equal to lOz+y.
Hence, by the question,
10x + y=3(a; + y);
/. lx=2y.
J^ROBLKMS. 175
Since x and y must both be positive integers not greater than 9,
it follows that z must be 2 and y must be 7. Thus the required
number is 27.
Ex. 7. The sum of a certain number and its square root is 90. What
is the number ?
Let X be the number ; then we have the equation
.. (x90)«=«,
or a;2181x + 8100=0,
that is (x 81) (a; 100) =0.
Hence «=:81, or »sslOO.
If, in the question, the square root means only the arithmetical
square root, 81 is the only number which satisfies the conditions.
If, however, * its square root * is taken to mean * one of its square
roots/ both 81 and 100 are admissible.
Ex. 8. The suih of the ages of a father and his son is 100 years \ also
onetenth of the product of their ages, in years, exceeds the father's
age by 180. How old are th^ ?
Let the father be x years old; then the son will be 100  x years
old. Hence, by the question,
^a;(100a;)=x+180;
.'. x»90x+ 1800=0,
that is («60)(a;30)=0.
Hence a;=60, or x=SO,
If the father is 60, the son will be 10060=40. If the father
is SO, the son will be 10030=70, which is impossible, since the
son cannot be older than the father.
Hence the father must be 60 and the son 40 years old.
Ex. 9. A man buys pigs, geese and ducks. If each of the geese had
cost a shilling less, one pig would have been worth as many geese as
each goose is actually wortii shillings. A goose is worth as much as
two ducks, and fourteen ducks are worth seven shillings more than
a pig. Find the price of a pig, a goose, and a duck respectively.
Let a;=the price in shillings of a pig,
y= .» »f •) ft goose,
and;E= „ ,, „ „ duck.
Then, by the question, a pig is worth y times (y1) shillings ;
.•. x=y(yl) (i).
176 i>roblems.
Since a goose is worth 2 dacks,
.*. 2^=22 (ii).
And, since 14 dacks are worth 7 shillings more than a pig,
liz=7+x (iii).
From (i) and (ii) we have the values of x and z in terms of y ; and,
substituting these values in (iii), we have
7y=7+y(yl),
or y«8y+7=0;
.•. y = 7, or y = l.
If y=7, x=42 from (i), and z = J from (ii).
If 1^ = 1, x=:0 from (i), and z=i from (ii). These values are how
ever inadmissible, since pigs cannot be bought for nothing.
Hence a pig cost 42«., a goose 7<., and a duck St. 6d,
EXAMPLES XV.
1. Divide 50 into two parts, such that twice one part is
equal to three times the other.
2. A has £6 less than B, G has as much as A and B
together, and A, B, G have £50 between them. How much
has eachi
3. One man is 70 and another is 45 years of age ; when
was the first twice as old as the second 1
4. How much are eggs a score, if a rise of 25 per cent, in
the price would make a difference of 40 in the number which
could be bought for a sovereign 1
6. A bag contains 50 coins which are worth £14 altogether.
A certain number of the coins are sovereigns, there are three
times as many halfsovereigns, and the rest are shiUings. Find
the number of each.
6. A can do a piece of work in 20 days, which B can do
in 12 days. A begins the work, but after a time B takes his
place, and the whole work is finished in 14 days from the
beginning. How long did A work 1
EXAMPLEa 177
7. A man buys a certain number of eggs at two a penny,
four times as many at 5d. a dozen, five times as many at Sd,
a score, and sells them at da. Sd. a hundred, gaining by the
transaction 3^. Qd. How many eggs did ho buy)
8. A bill of j£63. 5$. was paid in sovereigns and halfcrowns,
and the number of coins used was 100; how many sovereigns
were paid 1
9. A man walking from a town A to another B at the
rate of 4 miles an hour, starts one hour before a coach which
goes 12 miles an hour, and is picked up by the coach. On
arriving at B he observes that his coach journey lasted two
hours. Find the distance from A to B,
10. Two passengers have altogether 600 lbs. of luggage
and are charged for the excess above the weight allowed 3^. 4<a?.
and 1 1«. Sd, respectively. If all the luggage had belonged to
one person he would have been charged ^1. How much
luggage is each passenger allowed free of charge)
11. A piece of work can be done by A and Bin 4 days,
by A and (7 in 6 days, and by B and (7 in 12 days : find in
what time it would be done hj A, B and C working together.
12. A father's age is equal to those of his three children
together. In 9 years it will amount to those of the two eldest,
in 3 years after that to those of the eldest and youngest, and
in 3 years after that to those of the two youngest. Find their
present ages.
13. A and B start simultaneously from two towns to meet
one another : A travels 2 miles per hour faster than ^ and
they meet in 3 hours : if B had travelled one mile per hour
slower, and A at twothirds his previous pace they would have
met in 4 hours. Find the distance between the towns,
14. A traveller walks a certain distance : if he had gone
half a mile an hour faster, he would have walked it in ^ < .f the
time : if he had gone half a mile an hour slower he would have
been 2^ hours longer on the road. Find the distance.
S. A. 12
178 EXAMPLES.
15. Divide 243 into three parts such that onehalf of the
first, onethird of the second, and onefourth of the third part^
shall all be equal to one another.
16. A sum of money consisting of pounds and shillings
would be reduced to oneeighteenth of its original value if the
pounds were shillings, and the shillings pence. Shew that its
value would be increased in the ratio of 15 to 2 if the pounds
were fivepound notes, and the shillings pounds.
17. £1000 is divided between A, B, C and D. B gets
half as much as A^ the excess of CP& share over D'b share is
equal to onethird of ^'s share, and if ^'s share were increased
by J&IOO he would have as much as C and D have between
them ; find how much each gets.
18. Find two numbers, one of which is threefifths of the
other, so that the difierence of their squai*es may be equal to
16.
19. Find two numbers expressed by the same two digits
in different orders whose sum is equal to the square of the sum
of the two digits, and whose difference is equal to five times the
square of the smaller digit.
20. A man rode onethird of a journey at 10 miles per
hour, onethird more at 9 miles per hour, and the rest at 8
miles per hour. If he had ridden half the journey at 10 miles
per hour and the other half at 8 miles per hour, he would have
been half a minute longer on the journey. What distance did
he ride 1
21. Two bicyclists start at 12 o'clock, one from Cambridge
to Stortford and back, and the other from Stortf ord to Cambridge
and back. They meet at 3 o'clock for the second time, and they
are then 9 miles from Cambridge. The distance from Cambridge
to Stortford is 27 milea When and where did they meet for the
first timet
22. Divide £1015 among ^, J5, (7 so that B may receive
£5 less than A^ and G as many times 5*s share as there are
shillings in .i's share.
EXAMPLES. 179
23. On a certain road the telegraph posts are at equal
distances, and the number per mile is such that if there were
one less in each mile the interval between the posts would be
increased by 2^ yards. Find the number of posts in a mile.
24. The sum of two numbers multiplied by the greater is
144, and their difference multiplied by the less is 14: find
them.
25. A and JB start simultaneously from two towns and
meet after five hours ; if ^ had traveUed one mile per hour
faster and B had started one hour sooner, or if ^ had travelled
one mile per hour slower and A had started one hour later,
they would in either case have met at the same spot they
actually met at. What was the distance between the towns ?
26. A battalion of soldiers, when formed into a solid
square, present sixteen men fewer in the front than they do
when formed in a hollow square four deep. Required the
number of men.
27. A number of two digits is equal to seven times the
sum of the digits; shew that if the digits be reversed, the
number thus formed will be equal to four times the sum of the
digits.
28. A sets out to walk to a town 7 miles off, and £ starts
20 minutes afterwards to follow him. When B has overtaken
A he immediately turns back, and reaches the place from
which he started at the same instant that A reaches his
destination. Supposing B to have walked at the rate of 4
miles an hour : find A's rate.
29. A starts to bicycle from Cambridge to London, and B
at the same time from London to Cambridge, and they travel
uniformly : A reaches London 4 hours, and B reaches Cambridge
1 hour, after they have met on the road. How long did B take
to perform the journey 1
30. A number consists of 3 digits whose sum is 10. The
middle digit is equal to the sum of the other two ; and the
number will be increased by 99 if its digits be reversed. Find
the nnmber*
12—2
180 EXAMPLES.
31. Two vessels contain each a mixture of wine and water.
In the first vessel the quantity of wine is to the quantity of
water as 1 : 3, and in the second as 3 : 5. What quantity
must be taken from each in order to form a third mixture,
which shall contain 5 gallons of wine and 9 gallons of water 1
32. Supposing that it is now between 10 and 11 o'clock,
and that 6 minutes hence the minute hand of a watch will be
exactly opposite to the place where the hour hand was 3 minutes
ago : find the tima
33. A, B and C start from Cambridge, at 3, 4 and 5
o'clock respectively to walk, drive and ride respectively to
London. G overtakes B a,t 7 o'clock, and C overtakes 4 4 J
miles further on at halfpast seven. When and where will B
overtake A 1
34. A train 60 yards long passed another train 72 yards
long, which was travelling in the same direction on a parallel
line of rails, in 12 seconds. Had the slower train been
travelling half as fast again, it would have been passed in 24
HBconds. Find the rates at which the trains were travelling.
35. A distributes £180 in equal sums amongst a certain
number of people. B distributes the same sum but gives to
each person £Q more than A, and gives to 40 persons less than
A does. How much does A give to each person 1
36. Three vessels ply between the same two ports. The
first sails half a mile per hour faster than the second, and
makes the passage in an hour and a half less. The second
sails threequarters of a mile per hour ^ter than the third
and makes the passage in 2^ hours less. What is the distance
between the ports 1
37. Two persons 4, B walk from P to ^ and back. A
starts 1 hour after B, overtakes him 2 miles from Q, meets him
32 minutes afterwards, and arrives at F when fi is 4 miles off.
Find the distance from P to ^,
/.
CHAPTER XIL
Miscellaneous Theorems and Examples.
153. Elimination. When more equations are given
than are necessary to determine the values of the un
known quantities, the constants in the equations must be
connected by one or more relations, and it is often of
importance to determine these relations.
Since the relations required are not to contain any of
the unknown quantities, what we have to do is to eliminate
all the unknown quantities from the given system.
The following are some examples of Elimination :
Ex. 1. Eliminate z from the equations ox + & = 0, a'x + &' = 0.
From the first equation we have a;= — , and from the second
equation we haves 5;.
or
h V
Hence we most have  = ,, or ha' h'a=Oi which is the
a a
required resolt.
Ex. 2. Eliminate x and y from the equations
<Mf+&y + c=sO,
o'a; + 6'y + c'=0,
a"x + h"y+c"=Q.
182 ELIMINATION.
From the first two equations we have [Art. 143]
9i _ y _ 1
he* Vc co!  a' a a^a'h'
These yalaes of x and y most satisfy the third equation; henoe
ab'  ah ab'  ah
or a'*{fic'  Vc) + h'\(ia!  c'a)^d'{ab*  a'6) =0,
the required result.
The general case of .the elimination of n 1 unknown quantities
from n equations of the first degree will be considered in the Chapter
on Determinants.
Ex. 3. Eliminate x from the equations
As in Art. 143, we have
«» a? 1
hf^h'c" ccic'a ab'a'h"
Henoe {he'  Vc) (ah'  a'h) = (ca'  c'a)\
the required result.
It should be remarked that the above condition is also the
condition that the two expressions aa^^hx+c and a'a^ + h'x+c' may
have a common factor of the form xa; for if the expressions have
a common factor of the form xa they must both vanish for the
same value of x,
Ex. 4. Eliminate x from the equations
aV+6'a?+c'=0.
As in Ex. 3, we have
a^ X 1
7E»
be'  h'c ca'  c'a ah'  a'h
hc'h'e f ca'c'a y
•'• ab'^a'h~~\ah'a'h) '
/. {he'  Ve) {dV  a'h)'^= (ea'  c'a)\
the required relation.
ELIMINATION. 183
Ex. 5. Eliminate z from the equations
aa^ + bx+c^O (i),
a'x»+b'x^+e'x+d'=:0 (ii).
Multiply (i) by a'x, (ii) by a, and subtract ; then,
{db'ba')x^+(ac''ca')x+ad'=0 (iii).
We oan now eliminate x from (i) and (iii) as in Ex. 3.
Ex. 6. Eliminate x, y^z from the equations
x + y\z = a (i),
x'+y^ + z^=h^ (ii),
s^+y^ + z^=c^ (iii),
xyz—d^ (iv).
From (i) and (ii) we have
^z+2zx\2xy=d?'V^,
From (iii) and (iv) we have
a^+y^^z^Zxyzc^M^,
i c. {x\y\£^(x^+y'^+z^yzzxxy)=c^^d^.
Hence a{6*~ J(a2_ 6«)} =c3  Sd^;
.. a» + 2c»6d33a62=0,
the required result.
Ex. 7. Eliminate re, y, z from the equatioas
x^(y+z)=a? (i),
y3(2+x) = 6a (ii),
««(x+y)=c3..... (iii),
xyz=abc (iv).
From (i), (ii), (iii) by multiplication
« V«' (y +«)(«+ «) (« + y ) = a^^ W
Hence, from (iv),
(3f+z){z+x){x+y)=:l,
that is,
2xyz+x^{y+z)+y^(z+x)+z^(x+y)=l;
the required result.
Ex. 8. Eliminate I, m, n, T, m', n' from the equations
IV = a, mm' = 6, nn' = c,
wn' + m'n =r 2/, ni' + n'l = 2^ , int' + i'm = 2/i.
184 ELIMINATION.
By continned moltiplioation of the last three equations, we have
Sfgh = 2lmnVm;nf + W (mV« + mfhi^)
+ mm' {nH'^ + nf^l^ + rm' (Ihnf* + r%2)
= IV {mnf + m'n)^ + mm' {nV + n'Q*
+ nn' (Im' + Z'm)2  UVmm'nn'
Hence a6c + 2fgh ap hg^  ch^ = 0.
154. To find the condition thai the most general quad
ratic expression in x and y may he expressed as the product
of two factors of the first degree in x and y.
The most general quadratic expression in x and y may
be written in the form
a>x* + 2hxy + by* + 2gx {2fy + c (i).
What is required is the condition that the above ex
pression may be identically equal to
(lx + my + n)(l'x + m'y {n') (ii),
where I, m, n, l\ m\ n' do not contain x or y.
Now if (i) and (ii) are identically equal we may
equate the coefficients of the diflferent powers of x and
also of y [Art. 91]. Hence we have
IV = a, mm* = 6, nn' = c,
mw' 4 m'n = 2/, nr + n'l=^2g, Im' + l'm^2h
Eliminating Z, m, n, l\ m\ n [Art. 153, Ex. 8], we have
abo + 2fgh aP^hg^ ch^ = 0,
the condition required.
Ex. 1. For what value of X is
12a;«  lOajy + 2y2 + lla;  5y + X
the produot of two factors of the first degree in x and y?
Ans, Xss2.
Ex. 2. For what value of X is
12«^ + 36a5y + Xy' + 6x + 6y + 3
the product of two factors of the first degree in x and 2^ ?
An$, X=28.
EQUATIONS WITH RESTRICTIONS. 185
155. Equations in which there it some re
striction on the values of the letters. A single
equation which contains two or more unknown quantities
can be satisfied by an indefinite number of values of the
unknown quantities, provided that these values are not
in any way restricted. If however the values of the un
known quantities are subject to any restriction, a single
equation may be suflScient to determine more than one
unknown quantity.
For example, if we have the single equation 2a?+5y=7,
and restrict both x and y to positive integral values, the
equation can only be satisfied by one set of values, namely
by the values a? = 1, y = 1.
Again, jfrom the single equation
with the restriction that all the quantities must be real,
we can conclude both that a? — a = 0, and that y  6 = ;
for the squares of real quantities must be positive, and
the sum of two or more positive quantities cannot be zero
unless they are all zero.
Ex. 1. If (a+6+c)2=3(6c + ca+a5), then a=h==c.
We have a'+6a+c"6ccaa5=0,
that is J{(6c)2+(ca)3+(a~6)»}=0.
Whence 6  c, c  a and ah must all be zero.
Ex. 2, If X, 7i, 2/) 2^' be all real, and
then will x^vi and y^}f»
We have
a^(y^ + 2y'*2yy')2a!a^(y«+y'»S/y') + a/«(2y«+y''22^')=0;
.. (a!»2a»'+aJ^)(ya2yy'+y'2)+ary*2axB'yy'+«'V=0,
that is (a;«0'(yy')«+(a^'«'y)«=O.
Hence xy'xfy=0 and {X'x'){y y')=0.
From the second relation «=«' or y=y'; and either of these
combined with the first relation shews that both x=:xf and y=y\
186 IDENTITIES.
,Ex.3. If ai«+a,«+a,«+ =1^^
V+V+V+ =9^
and Oi&i + a^j+a,6,+ =M»
the quantities being aU real; then will
61 6, ftj ^
Multiply the equations in order by j*, t^ and  2pg respectively,
and add ; we then have
(gaii)6i)* + (5a,p6a)«+(ga,jp6,)«+ =0.
Henoe goj 2J&i=0=gaj pftjssgo, p68= <fec.
Therefore ^=^ = ^ = ^= &c.
156. We have already proved that
= i(a + 6 + c)((6c)' + (ca)* + (a6y}
= (a + 6 + c) (a + w6 + (o\){a 4 a)'6 + wc),
where © is either of the cube roots of unity. [See Art.
139.]
From the above many other identities can be found.
Ex. 1. (6 + c)» + (c+a)« + (o + 6)»8(6 + c)(c + a)(a + 6)
=2(a»+6» + c»8a6c).
Left side = i{6 + c+c+a+a + 6} {(c + aa + 6)*+ two similar
terms}
= (a+6+c){(6c)a+(ca)»+(a6)»}
=2(a» + 6» + c»3a6c).
Ex.2. (6+ca)»+(c + a6)» + (a+6c)»
8(6+ca)(c + a6)(a+6c)=4(a»+6'+c»8a&c).
Left side = j^ (a + 5 + c) { (2d  2c)3 + two similar terms}
=4(a*+6»+c»3a6c).
Ex.8. (a?yzf{{3^''Zxf^'(z^'XyY^(x^yz)(y^'Zx)(}^'QBy)
Left side = J(a^+y'+«'y««Ba^[(y*£x«^a;y)' + two
similar terms]
= i(«'+y'+«'y««5«y)(«+y+«)'[(y«)' + t^o
similar terms]
= (x+y + z)^{x^\y^+z^yzzxxy)^
EXAMPLES. 187
Ex.4. Shew that (a» + y» + x;" 3ajy2)(a»+6»+c» 8a6c) can be
expresBed in the form Z»+ T^kZ^ZXYZ.
We have
(«+y+«)(a + 6+c) = (aa;+6y+c>;) + (&»+cy+ajB) + (ca;+ay+62),
(« + «y + fci"^;) (a + w^ft + wc) = (oaf + 6y + c;?) + «* (6« + cy + a«)
and
(j; + ci)^ + w«) (a + (tf& + (tf'c) = (oo; + &y + C£) + w (&dB + cy + 02)
The continned product of the left members of the above equations
is
and the continued product of the expressions on the right is
(005 + 6y + c«)' + (feu + cy + o^)* + (ex + oy + 6«)'
 3 (ax +hy\ez) (hx kcy+az) (ex + ay + hz) ,
which is of the required form.
157. Definitions. The symbol = is often used to
denote that the two expressions between which it is placed
are identicaUy equal Thus a* — 6" = (a + b) {a — 6).
The sum of any number of quantities of the same
type is often expressed by writing only one of the tenns
preceded by the symbol 2. Thus Xbc means the sum
of all such terms as &o ; so that if there are three letters
a, b, c, Xbc ^bc{ca + ab. So also the identity
(a + 6 + c+...)' = a' + 6' + c*+...+ 2(a6 + 6c+...),
may be written (2a)' = 2a' + 22a6.
The product of any number of quantities of the same
type is often expressed by writing only one of the factors
preceded by the symbol 11. Thus 11 (6 + c) means , the
product of all such factors as (6 + c) ; so that if there are
three letters a, b, c, 11 (6 + c) = (6 + c) (c + a) (a + 6).
168. The following examples illustrate cases of fre
quent occurrence.
188 EXAMPLEa
Ex. 1. H a»+6»+c»=(a+6+c)«, then y/^
where n is any positive integer.
Since (a + 6 + c)» = a> + 6»+c'+3(6 + c) (c+a)(a+6), the given
relation shews that (6 +c){c+ a) (a + 6) = 0.
Hence either 6+6=0, or e+a=0 or a+&=0.
K 6 + c = 0; then fe^^+i = (  c)*»+i =  c«»+^ and therefore
6a»Hi+c*»+i=0.
Thus if 6 + c = 0, a*^* + 6*^i + c*^i(a+6+c)2»+i becomes
^2n+l ^ jai^f 1 + c^Hirl _ ^21141  52H+1 ^. c*^i = 0.
Hence a«»+i + 6*»+i + c*»+^=(a + 6 + c)*»+i if &+c=0; and so also
ifc + a = 0, or if a + bssO, This proves the proposition, since
(ft + c)(c + a)(a+6)=0.
Ex. 2. U X, y, g he unequal, and if
prove that each equals 2xyZt and that x+y+z + m=0.
Since y' + «'+m(y'+«*)ss«»+jc»+i»(t?+a^), we have
that is (ya5){"y2^a!y + a:2+m(a; + y)}=0.
Therefore, yx not being equal to zero, we have
y^+xy+{t^+m(x+y)=0 (i).
So also, since y^z,
«' + y«+y'+m {z+y)=0 (ii).
From (i) and (ii) we have by subtraction
sfiz*+y{X'z)+m(xz)=0.
Hence, as jb4s«, we have
«+y+«+m=0 (iii).
Substitnte  (x+y +£) for m in (i) ; and we have
a^+«y+y'(«+y)(a; + y + «) = 0;
/. yz + zx^xy=0 (iv).
Then y^'hz^ + m{y^^z^=y9+s^(y^+z^{x+y+z) from (iii)
=  (Sf^x+z^+yh+zhi)
= y {xy+yz)z (yz+zx)
= 2xyz from (iv).
EXAMPLES. 189
Ex. 8. Shew that, if a+b + c+d=0, then will
a* + 6* + c*+(i*=2(a6cd)« + 2(acM)« + 2(ad6c)2 + 4a6cd.
We have to prove that
2a4=22aa6»8a6cd.
Since a + b+c + d=0; we have, by squaring and transposing,
a^+l^'\c^'\cP=2{bc + ca + ab + ad+bdhcd).
Hence by squaring
2a* + 22a«6«=4(S6c)«.
Now (S6c)«=26»c»+6a6cd+26cd(6+c + d) + 2cda(c+<f+a)
+ 2da6(d+a+6)+2a6c(a+6 + c) = S62c3 + 6a6cd8a6cd.
Hence Za* + 22a«6«= 42a26»  8abed ;
/. 2a*=2Sa«6a8a6cd.
Ex.4. Prove that, if ax+by + czszO, and H h = 0, then will
* ' X y z '
a3fi+by^+cz^= '{a+b + c)(y+z)(z+x)(x+y).
From the given relations we have, as in Art. 143,
a b c
y z " z X ^ X y '
z y X z y X
Hence [Art. 113] each fraction is equal to
aat^kby^+ez^ a\b+c
\z y) ^ \x zj \y xj z y X z y X
Hence
a^bVc ~ x{:y^ ~ z^){y (z*  x^)\' z(3? y^)
_ {y^z^){z^x^)(x^y^)
(yz)(zx){xy)
= (y + z){z + x) (x + y).
EXAMPLES XVL
1. Shew that, if = a, = b and = c ; then
y + z z +05 x^y
.„ <* h c 
will + r — = + = = 1.
1 +a 1+6 1 +c
190 EXAMPLES.
2. Shew that, if ooj + 6y = and cas* + dxy + ey* = 0, then
will a*e + 6*c = abd,
3. Eliminate x, y, z from the equations
y + z z + x x + y ^ /w^>M.x^
4. Eliminate x, y, z from the equations
y X z y , X z
a; i2; y x z y
1 1 1
5. If a? +  = 1 and v +  = 1 ; prove that z + =l,
y ^ z ^ X
6. Eliminate x from the equations
b ,
a ) c = 005,
X
a — c = — ox,
X
7. Eliminate x, y, z from the equations
x' — yz = af y' — zx^^bf s^~xy=Cj ax i by + cz = d.
8. Prove that the equations
x + y + z = a,
a5* + y" + »•  3osyz = c\
do not give any roots, but simply a relation between a, b and c.
9. Shew that, if
bz+cy = cx + az = ay + bXf and aj* + y* + «"2y22«a; 2ajy = 0;
then will a * 6 ± c = 0.
10. Shew that, if  +^ + = 1 and  +  + =0; then
a b c X y z
a? y* »■
will ^ + Tr + T = l«
a b <r
111
11. If a5 +  = y +  = » + ; then 0*^21* = 1, or a5 = v=«.
EXAMPLES. 191
12. Shew that, if x^cy^bz, y = az\cx and z = bx^ai/ ;
then
a^ y ^
l»a'"l6»""lc''
13. Shew that, if ic* = y* + «" + 2ai/Zf y^ = z* + a:^ + 2bzx and
z^ = af +^^ + 20X1^; then
^ __ y* _ *'
14. Shew that, if x, y, zhe unequal, and
a + bz a + bx , a + by
y = J, z = 3andaj= jF^ ,
c+ dz c + oa? c^dy
then will adhbc + 6* +©■ = 0.
15. Eliminate a?, y, « from the equations
yz or zx y* xy z
16. Eliminate a:, y, «; from the equations &a^ + Zx* + c = 0,
cy'+my + asO, a»' + w« + 6 = 0, xyz^l.
17. Eliminate a?, y, « from the equations
i^ ^ n^ z= ayZf js" + 05* = 6«a5, x^ + f^^cxy,
xyz not being zero.
18. Eliminate (i) as, y, 2; and (ii) a, 5, c from the equations
1 y z z X , i ^ 1 y
6 + c=a, c + a = o, and a  + 6  = a
» y aj « y as
19. Eliminate as, y, ^2; from the equations
ax + yz = bCf by + zx = ca, cz¥xy = ab, and xyz = <i6c.
20. Eliminate a;, y, z from the equations
a^ — xyocz _^f^ — yz'yx ^z^ — zx — zy
a ~ 6 "" c '
and oas + 6y + c« = 0.
192 EXAMPLES.
21. From the equations €^y% = a'{y + «)', b'zx = j8* (« + a;)",
c'xy = y* (a; + y)', deduce the relation
abc _€? 6* ^ A
22. Prove that, if
and yz + zx + xy = 0; then will a ± 6 ± c = 0.
23. Prove that, if  + + = r , then will
a c (a\ h '\ c)
1.
where n is any positive integer.
24. Shew that, if
2bc ■*■ 2ca '^ 2ab '
then (6 + c  a) (c + a  5) (a + 6  c) = 0,
and
/ ^Vc'a V'^^ /c'+a'by ^' /a'hb'cy "''_
\ 25c 7 "*" \ 2ca y ■*" V 2a6 /
25. If a V + by + c V = 0,
aV + 5y + c V = 0,
and — a' = — 6' = — c':
X y z
prove that aV + 6y + cV = 0,
and a V + 6y 4 c V = a V + b*y* + c V.
26. If a;^ = y^ = »^, and a;, y, 2; be unequal;
XT jf z
then each member of the equations is equal toa; + y + « — a.
iz  a;)*
27. If aj, y, 2J be unequal, and if 2a3y = ^ ~ and
2a  3« = fc:i^ , then will 2a 3a;= ^^Ll^ , and
a a;
aj + y + ;5 = a.
EXAMPLES. 193
28. If 35+ Hi — 5 g be not altered in value by inter
XT + y 4 Z
changing x and y, it will not be altered by interchanging x and
«, and it will vanish if a5 + y + «=l, the letters being all
unequal.
29. If Xf y^ z be unequal, and
y' + «" + m (y + «) = 2j" + oj' + m (« + a:) = a' + ^ + m (a; + y ),
then each will equal 2xyz,
30. If X, y,zhe unequal, and
. y* + 2* + myz = «* + 03* + mzx = af + y' + mocy,
then each will equal ^(05* + ^* + s^).
31. If X, y be unequal, and if (^^LZlZll = (2y«^' ^
then will each equal ^ — .
z
32. Shew that, if a, 6, c, c? be all real quantities not zero,
and {a^ + 5*) (c* + c?') = 4a6cc? : then will a = ± 6 and c=^d.
33. If a, 5, c, 0? be all real quantities, and
(a» + 6»)ar"26(a + c)a; + 6' + c" = 0;
then  =  = a;.
6 a
34. Shew that, if
(aj* + y + »*) (a* + 6* + c') = (ax + hy + cz)*,
then aj/a = y/b = »/c.
35. Prove the following :
(i) If 2 (a' + b') = (a + b)', then a = 6.
(ii) If 3 {a' + b' + c') = {a + b + c)% then a=: b =c.
(iii) If 4 (a* + 6' + c* + c?') = (a + 6 + c + c?)*, then
a=:b = c = d.
and
(iv) If w(a' + 6' + c*+ ) = {a + b + c+ )*, then
a=ib = c = , n being the number of the letters.
S. A. 13
194 EXAMPLES.
36. Prove that, if a, b, c, d be all real and positive, and
a* + 6* + c* + c^ = iabcd;
then will a = b = c=^d,
37. If
(n 1) x' + 2x(a^  aj +a,'+ 2a/ + 2a3*+ ... + 2a'.i + aj
for real values of as, a^ a^, ..., a^ ; then will
Verify the following identities :
38. a' (6 + c) + b' (c + a) + c' (a + 6) + a6c (a + 6 + c)
= (a* + 6' + c') (5c + ca + a5).
39. (b + c^a'dy{b^c){ad)+(c+abdy{C'a)(bd)
+ (a + 6  c  c?)* (a  6) (c  c?)
= 16 (5 c) (c a) (a6) ((^a) (c/5) ((£c).
40. 8 (a + 6 + c)» (6 + c)»  (c + of  (a+ 6)»
= 3 (2a + 6 + c) (a + 26 + c) (a + 6 + 2c).
41. {a+b + c + d)'^'{b + c^dy{c + d + ay{d+a + bf
 (a + 6 + c)' + (6 + c)* + (c + ay + (ahby + (a + cQ*
+ (6 + c?)» + (c + c?)*a»  6* c* ef = eOabcd (a+b + chd),
42. (a + 6 + c)' abc(bc + ca^ aby = a5c (a* + 5' + c')
(6V + cV+aW).
43. (a* + 6» + c«)' + 2 (6c + ca + a6)«
 3 (a' ^b^ + c')(bc+ ca +aby ^ia"" + 6» + c»  3a5c)«.
44. (cab') {abc') k {ab c') {be a') ■^{bc'~a')(cab')
= (6c + ca + a6) {be + ca + ab a^ b^ c*).
EXAMPLSa 195
45. 2(c" + ca + a»)(a« + a6 + 6')(6' + 6c + c7
+ 2 (a» + a6 + 5") (6* + bc+ c')  (c» + ca + a')'
+ 2(6' + 6c + c')(c« + ca + a')(a» + a6 + 67
= 3 (6c + ca + oft)'.
46. Shew that
{Sab  cy + (Sbc ay + {3c a hf
3(3a5c)(36ca)(3ca6)=16(a' + 6'' + c»3a6c).
47. Shew that
(7ia~ 6 (?)• + (w6 c ■a)' + (rwja 6)*
— 3 (na — 6 — c) (w6 — c — a) (?ic — a  6)
= (n+iy (ii2) (a» + 6' + c'Sabc).
48. Shew that
(«» + 2yzy + (y» + 2zxy + («" + 2a^)»
 3 (a" + 23^«) (y' + 2«a;) (is" + 2a:y) = (a?' +/ + »• 3a;2/2;)'.
49. Shew that
(by + rts:)' + {bz + oo:)' + (6a; + ay)'  3 (6y + o«) (6« + <ix) (bx + ay)
= (a'+ 6*) (a;* + y" + »* 3a;y2).
50. Shew that, if 1 + w + w' = 0, then
[(b  c) (a; a) + 0) (c a) (a;  6) + a)' (a  b) (« c)]'
+ [(6  c) (aj r a) + CD* (c  a) (aj  6) + CD (a  6) (a;  c)]'
= 27 (6  c) (c a) (a  6) (a?  a) (a?  6) (a? c).
51. Shew that the product of any number of factors, each
of which is the sum of two squares, can be expressed as the
sum of two squares.
52. Verify the identity
(a* + 6' + c' + c?*) (^' + 5^ + r*+ «■) = (a;? + 63' + cr + di)^
+ (aq  6p + c* — dry+ (or ^bscp + dqy
■{'{ixs + br'Cq d/py.
Hence shew that the product of any number of factors,
each of which is the sum of four squares, can be expressed as
the sum of four squares.
13—2
196 EXAMPLES.
53. Shew that (aj^+ xy + ^) (a* + a6 + 6*) can be expressed
in the form X' + XY^ Y\
54. Shew that (of •{■ pxy \ qy^) {a' + pab + qb^) can be ex
pressed in the form X' ^pXYiqY',
55. Shew that, if 2s = a + 6 + c,
(i) a («  6) (s  c) + 6 («  c) (s  a) + c (s  a) (s  b)
+ 2 (sa) {8 — b){s c) = abc.
(ii) (8  ay + (« ^ 5)« + (s  c)" + 3a5c = 8\
(iii) (6 + c) s («  a) + a (s — 6) (s — c) — 2s5c
= (c + a) s (s — 6) + 6 (s  c) (s — a) — 2sca
a= (a + 5) 5 («  c) + c (« — a) (s — 6)  2sa6.
(iv) a(6 c) {8ay + 6 (c a) (5  6)»
(v) « (s  5) (s  c) + s (s  c) (s  a) + « (s  a) (s  6)
(««)(« — 6) (s  c) = abc,
(vi) (s  a)« (8  6)« («  cy + s' (5  5)' (s  cy
^s' {8 cy {8ay + 8' {8  ay (8 by
+ s («a) (5  5) («  c) (a' +b' + c^ = a'b^cK
56. Shew that, if 2s = a + 6 + c + c?,
Shew also that
a{8  b)(8 c)(8  d) + b{8  c) (s d)(sa) + c{8d) (sa) (8b)
^d(8a)(8b){8c) + 2(8a)(8b){8c){sd)
 8 (bed + cda + dab + abc) =  2abcd.
57. Shew that, ifa+6 + c+cf = 0, then
ad(a + dy + bc{ady+ab(a + by + cd{aby
+ ac{a + cy\ bd {a  cy + iabcd = 0.
EXAMPLES. 197
58. Shew that, if
{a + 6) (6 + c) (c + d) {d^a)
= {a + h + c + d) (bed + cda + dab + ahc) ;
then ac = bd.
59. Shew that, if a + 6 + c = and a; + y + 2; = 0, then
4 (005 + by + c%y  3 (005 + 63/ + cz) {a' + 6' + c") (05* + 2/' + 2^)
 2 (5  c) (c  a) (a ~ b) {y z)(z x)(x  y) = 5iabcxyz,
60. Shew that, ifa + 6»c = 0; then
(i) 2 (a' + 6' + c^) = 7a5c (a* + 6* + c*).
(ii) 6(a^ + 6^ + c0 = 7(a'' + 5" + c»)(a*+6* + c^).
(iii) a' + b^ + c^ = 3a*6 V + J (a' + 6« + c^.
(iv) 25 (a' + b' + c') (a' + b' + c«) = 21 (a* + 6* + c^)^
61. If a + 6 + c + c?= 0, prove that
(a^ + 6' + c' + (f )' = 9 (bcd+cda + cM + a5c)*
= 9 (be — ad) (ca — 5cZ) (ab  cd),
62. Shew that, if a + 6 + c = 0, then
(6c c — a a — bX/ a b ^\_n
a 6 c J\bc ca a—bj
63. Prove that, if
1 tn Tim
\ +l\ln 1+m + m^ l+r» + ?im
1,
 I ml 1 
and = — = — =— + :: = + z = 1,
and none of the denominators be zero, then will l — m^n.
64. Shew that
a + (1  a) 6 + (1  a) (1  5) c + (1  a) (1  6) (1  c) (^
+ ... = l(la)(l6)(lc)(lc/) ..
198 £XAMPLEa
65. Shew that
l=l+2(la) + 3(la)(l2a) + ...
+ {«(! a) (1  2a) ... (1  n  la)}
+ l{(la){l2a)...(lna)}.
66. Shew that
a" + a'(la") + a»(la")(la"^) + ...
+ {a(la)(la»)...(la')} + {(la)(la^)...(la)}
= 1.
67. Shew that, if w be any positive integer,
lg" {\ar){\ar') (1 ^(l a*" *) (1a"^
1 —a
68. Prove that, if
a + h + c + d=Oy
and oa; + 6y + C2J + c?t4 = ;
then 2 (a^oj + 6*y + c^z + c?*^)
= (a*a; + h*y + c*2; + d*u)(a' + 6" + c* + cP).
69. Prove that, if 7i be any positive integer,
1111 111 1
2n 71+1 n + 2 ' *" 2/i
70. Prove that, if
u V u + a V b^ u+a' v  b' ~^ / '
then /^ (ab'  a'bf = aa'bb' {a  a') {b  b').
/^CHAPTER XIIL
Powers and Roots. Fractional and Negative
Indices.
159. The process by which the powers of quantities
are obtained is often called involution; and the inverse
process, namely that by which the roots of quantities are
obtained, is called evolution.
We proceed to consider some cases of involution and
of evolution.
160. Index Laws. We have proved in Art. 31, that
when m and n are any positive integers,
oTxar^dr^ (i).
This result is called the Index Law.
From the Index Law, we have
and so on, however many factors there may be.
Hence aTxaTxaF y. ...=a™"^"*^ (ii).
Thus the index of the product of any number of powers
of the same quantity is the sum of the indices of the factors.
Also, a*" X a"* X a*" X ... to w factors
= ^'»»+»»+»*+ to n terms
200 POWERS.
Hence (ary = ar'* (iii).
Thus, to raise any power of a quantity to any other
power, its original index must he mvltiplied by the index of
the power to which it is to he raised.
Again, to find (at)*.
(a6)"* = a6 xah x abx torn factors, by definition,
= (a X a X a to m factors) x(bxbxb to
m factors), by the Commutative Law,
= a** X ft*", by definition.
Hence (aby = a"* x ft*
Similarly (abc. . .)*" = a** x 6"* x c"* x (iv).
Thus, the mth power of a product is the product of Hie
mth powers of its factors.
The most general case of a monomial expression is
a'Vc'
Now (a'Vcf )*" = (a*)"* (&'')"• {c^ from (iv)
= a!^h^d^ from (iii).
Hence (a*tV )"» = a'^fc^c'". . . ...(v).
Thus any power of an expression is obtained by taking
each of its factors to a power whose index is the product of
its original index and the index of the power to which the
whole expression is to be raised.
As a particular case
i^X^ia X ^Xax(^T^a x 1 =^
\b) V 6/ " \b) ""b b'
161. It follows from the Law of Signs that all powers
of a positive quantity are positive, but that successive
powers of a negative quantity are alternately positive and
negative. For we have
ROOTS OF ARITHMETICAL NUMBERS. 201
(a)« = (~a) (a) = ^a^
(a)» = ( of (a) = (+a^ (^a) = 'a\
and so on.
Thus ( a)** = + a^, and ( af""^' =  a*»^\
Hence all even powers, whether of positive or of
negative quantities, are positive ; and all odd powers of
amy quantity have the same sign as the original quantity,
162. Boots of Arithmetical numbers. The
approximate value of the square or of any other root of
an arithmetical number can always be found: this we
proceed to prove. It will be seen that the process
described would be an extremely laborious one; we are
not however here concerned with the actual calculation
of surds.
Consider, for example, v^. First write down the
squares of the numbers 1, 2, 3, &c. until one is found
which is greater than 62 : it will then be seen that 7' is
less and 8' is greater than 62. Now write down the
squares of the numbers 7*1, 7'2, 7*3, ..., 7*9: it will then
be seen that (7'8)* is less, and (7*9)* greater than 62.
Now write down the squares of 7'81, 7*82, ..., 7'89 : it will
then be seen that (7 '83)* is less, and (7*84)' greater
than 62.
By continuing this process, we get at every stage two
numbers such that 62 is intermediate between their
squai*es, and such that their difference becomes smaller
and smaller at every successive stage; moreover, this
difference can, by sufficiently continuing the process, be
made less than any a^ssigned quantity however small.
Thus, although we can never find any number whose
square is exactly equal to 62, we can find two numbers
whose squares are the one greater and the^other less than
62, and whose difference is less than'^any assigned quantity
however small. The limiting value of tfiese two numbers,
202 SURDS OBEY FUNDAMENTAL LAWS.
when the process is continued indefinitely, is called the
square root of 62.
The process above described for finding a square root
can clearly be applied to find any other root.
Thus an nth root of any integral or fractional number
can always be found.
163. Surds obey the Fundamental Laws of
Algebra. The fundamental laws of Algebra were proved
for integral or fractional values of the letters ; and it can
be proved that they are also true for surds.
Consider, for example, the Commutative Law.
We have to prove that
We can find whole numbers or fractions x, y and p, q
such that
and p>'!^b> q;
and the difference between oc and y, and also the difference
between p and q, can be made less than any assigned
quantity however small.
Hence oo xp> ^^a x"ijb>y xq,
and pxx>'^b xj^a>qxy,^
But, since oc, y, p, q are integral or fractional numbers,
we know that x xp=p xx, and yxq = qxy; also the
difference between px and qy can be made less than any
assigned quantity however small.
It therefore follows that ^a x XJh and ^6 x ^J^a, which
are both always intermediate to xp and yq, must be equal.
Thus the Commutative Law holds for Surds, and the
other laws can be proved in a similar manner.
164. We already know that there are two square
roots, and three cube roots of every quantity ; and we may
remark that there are always n nth roots. Thus there is
^ PROPEBTIES OF ROOTS. 203
an important diflference between powers and roots; for
there is only one nth power, but there is more than one
nth root.
165. We have proved in Art. 160 that the mth power
of a product is the product of the mth powers of its factors;
and, since surds obey the fundamental laws of Algebra, the
proposition holds good when all or any of the factors are
irrational. Hence
{fjax f^b..,y = (\/ay X (<s/by ... = ab....
Also {Jab...y = at. . . , by definition.
.'. {/Ja X *Jb...y= (^/a6...)^
Hence aJg x V6... must be equal to one of the square
roots of oJ... .
We can write this
\/a \/6. . . = ^ab. . . ,
meaning thereby that the continued product of either of
the square roots of a, either of the square roots of b, &c. is
equal to one or other of the square roots of at ...
Similarly we have, with a corresponding limitation,
;!^a "/a.
;/a76... = 7a6..., and Hb"^ \/b'
Also v^a*" = *t^a"*, for their wpth powers are both equal
to a"*".
Again, since the nth power of a monomial expression is
obtained by multiplying the index of each of its factors by
n, it follows conversely that an nth. root of a monomial
expression is obtained by dividing the index of each of its
factors by w, provided the division can be performed.
Thus one value of v^a* is a', one value of IJa^ 6® c' is
a* 6' c, and one value of Ha^ b^ c"^ is a* b^ c\
204 EXTENSION OF MEANING OF INDEX.
Fractional and Negative Indices.
166. We have hitherto supposed that an index was
always a positive integer ; and this is necessarily the case
so long as we retain the definition of Art. 9 ; for, with that
definition, such expressions as c? and a~* have no meaning
whatever.
We might extend the meaning of an index by assign
ing meanings to a" when n is fractional and negative.
It is, however, essential that algebraical symbols should
always obey the same laws whatever their values may be ;
we therefore do not begin by assigning any meaning to a"
when n is not a positive integer, but we first impose the
restriction that the meaning of a** must in all cases be such
that the fundamental index law, namely
shall always he true ; and it will be found that the above
restriction is of itself sufficient to define the meaning of a*
in all cases, so that there is no further freedom of choice.
For example, to find the meaning of a .
Since the meaning is to be consistent with the Index
Law, we must have
a*xa* = a*'^* = ai = a.
Thus a^ must be such that its square is a, that is cr
must be s/a.
Again, to find the meaning of a"\
By the index law
a'' xa' = a''^' = a^ ; therefore a'' = , = ,
a^ a
Thus a~* must be  .
a
FRACTIONAL AND NEGATIVE INDICES. 205
167. We now proceed to consider the most general
cases.
1
I. To find the meaning of a", where n is any positive
iuteger.
By the index law,
111
a" X d^ xa"* X to n factors
++1+ to « terms 
= a^ ^ ^ =0^ = 0^ = a,
1
Hence a" must be such that its nth power is a, that is
1
m
II. To find the meaning of a**, where m and n are any
positive integers.
By the index law,
a^xa^x to ri factors = a** » =a^ =a,
m
Hence a" = Ua"".
We have also
11 1,1... m
  . /. . ++...... torn terms —
a^ xa^x to m factors = a** ** = a**.
m 1
Hence a''={arf.
Thus we may consider that a* is an wth root of the
mth power of a, or that it is the mth power of an nth root
of a ; which we express by
With the above meaning of a** it follows from Art.
m mp
165 that a" = a"^.
206 FRACTIONAL AND NEGATIVE INDICES.
Note. It should be remarked that it is not strictly
true that J^{ar) = {IJaY except with a limitation corre
sponding to that of Art. 165, or unless by the wth root of
a quantity is meant only the arithmetical root For
example, }/{a^) has two values, namely ±a', whereas ( ilaf
has only the value + a*.
III. To find the meaning of cP.
By the index law
a''xa'^ = a'^ = a'^; .. a* = a'"Ta'"= 1.
Thus a® = 1, whatever a may be.
IV. To find the meaning of a~™, where m has any
positive value.
By the index law,
a"*" X a"* = a"^"^ = a® ; and a® = 1, by IIL
Hence a^ = =. , and a"' = ^z, .
a a
168. We have in the preceding Article found that in
order that the fundamental index law, a"* x a"= a"***, may
always be obeyed, a"" must have a definite meaning when
n has any given positive or negative value. We have now
to shew that, with the Toeanings ihvjS obtained,
a"* X a" = a"*^, (a"*)" = a*"% and {ahf = aV,
are true for all values of m and n. When these have been
proved, the final result of Art. 160 is easily seen to be true
in all cases.
INDEX LAWS. 207
I. To prove that a*^ xa'*^ a"*"^", for all values of m
and n.
We already know that this is true when m and n are
positive integers. Let m and n be any positive fractions ~
and  respectively. Then
a"* X a" = a* X a' = ^a^ x ^^a**. by definition
= ^^ X V^= "iJar^ [Art 165]
p»+rq
s= a «' , by definition
Thus the proposition is true for all positive values of
m and n. To shew that it is true also for negative values,
it is necessary and sufficient to prove that
a"~ X a~* = a"~"*, and a"* x a"" = a"*"*
where m and n are positive.
Ill
Now a"^ X a~* = — X — = ^, = a"^"".
a a a
And, if m — n be positive,
a"*"* X a" = a"*, and a"* x a"* x a" = a** ;
therefore a*""* = a^ x a
.~»
Hence, if m  n be negative, ^x^—,,,
that is, a"^ xa* = a"*"".
Hence a"* x a* = a"*^, for aZi values of m and w.
Cor. Since a"**" xa^=^a!^ for all values of m and n,
it follows that aT^oT^ a'
.m— n
208 INDEX LAWS.
II. To prove that (a*")* = a*"*, for all values of m and n.
First, let n be a positive integer, m having any value
whatever.
Then {ary=^ar xoTx a"' x to n factors,
^ Qin+m\m\ to n terms "Jjy J^
= »*"".
Next, let w be a positive fraction  , where p and q are
positive integers.
Then (a'^)'' = (a'7=y{(aT}, = VK"), since p is an
integer,
mp
= a* =a"'".
Finally, let n be negative, and equal to —p.
1 1
Then (a)" = (a"") = ^^ = _ = a"^ = a"".
Hence for all values of m and ti we have
III. To prove that (a6)* = a"6*, for all values of w.
We have proved in Art. 160 that (aby = a%\ where
w is a positive integer.
And, whatever m may be, provided that q^ is a positive
integer, we have
^a^iy = a^"^ X dr}r X ... to 5 factors
;_ ^w+m+... to g terms ^ ^»»+»»+ „, to g terms
Let n be a positive fraction , where p and g are
positive integers. Then
RATIONALIZING FACTORS. 209
(aby=(ahy=^X/(aby=;^(a''b^), since p is a positive
integer.
Also (a"6"/ = a**^6"*, since g is a positive integer.
Hence a"6" = i/((^If) = {aby.
Thus (aby = a^t*, for all positive values of n.
Finally, if n be negative, Bud equal to — w, we have
(aby = (ab)'^ = T^ = —4 = a""*6"^ = a"6*.
Ex. (i) Simplify a^xa^K
Ex. (ii). Simplify aH^xaH^,
Ex. (iii). SimpHfy (a^b^) " ^.
(a«6^)t=a(«)(t)6*(«=a36»=^'
Ex. (iv). SimpUfy \/(a*6«c^)fAy(a^Mci).
169. Rationalizing Factors. It is sometimes re
quired to find an expression which when multiplied by
a given irrational expression will give a rational product.
The following are examples of rationalizing factors.
Since (a+Jb){aJb)=a^b, it follows that a^Jb is made
rational by maUiplying by aw,Jb,
So also a^Jb :i^Cy/dia made rationtil by multiplying hja^b^e ijd.
Again from the known identity
= (a+6 + c)(a + 6 + c)(a6 + c)(a + 6c),
S. A. 14
210 RATIONALIZING FACTORS.
it follows that the rationalizing factor of
^p+^q+^riB(Jp\s/q¥^r)(^p'^q\s/r)(Jp + Jq"Jr).
The rationalizing factor of s/p+jq + ijr may also be found as
follows,
{s/p+>Jq+^r){Jp+^qs/r)=p + q''r\2jpqf
and
(p+gr+2,^)(p + gr2^/pg)=(p + gr)24i?g.
Thus the required rationalizing factor is
UP+Jq's/r){p + qr2jpq),
which is the same as before.
Again » from the identity
(a + 6)(a2a5 + 6«)=a8 + 6',
the rationalizing factor of a+b^ is seen to be a^  ab^ + bs,
170. To find the rationalizing factor of any binomial.
P r
Let the expression to be rationalized be cw?*± by'.
P T.
Put X= ax*, and T=by\ and let n be the L.CJtf. of
q and s.
Then it is easily seen that X" and F" are both
rational.
Hence, from the identities
and (XF)(Z'*^ + Z"*F+ + F«^) = Z" F*,
the rationalizing factors ofX+F and X— F are seen to
be respectively
Z'*'X"«F+ + ( i)«^F'*"^
and X^ + S:""'F+ + Y*'\
EXAMPLES. 211
Ex. To find a factor which will rationalize
Here X=x^, F=ay^, n=6.
The fiftctor required is therefore
EXAMPLES XVIL
1. Simplify a^b^ x a~H~\
2. Simplify a'^ x a'i x (a')"* x — .
3. Simplify (a^'V)* x (a«5»c"j*.
X il^.Ji c+a J_ a+b J_
4. Simplify {^aja^b ^ (ajaftjbc X (aj^^)ca .
5. Multiply x^ + x^y^ + y^ by x^y^.
6. Multiply aj' + 1 + a?' by aj»  1 + «;«.
7. Multiply
8. Divide x^^ + x"^ by a^a;"^.
9. Divide a^x by a^^a;^.
10. Divide x^  xy^ + a^y ^ y^ by x^y\
11. Shew that
oj*  4a;* + 2aj* + 4a;  4aj^ + x^ = (a;*  2x^ + a;^)*.
12. Multiply 4a;'  5a;  4  7a;"* + 6a;"' by 3a;  4 + 2aj"»
and divide the product by 3a; 10 + lOa;"*  4a;~'.
14—2
212 EXAMPLES.
13. Divide
oj  a?"^ — 2 (o;^ — 05"') + 2 (a;« — a5~«) by as' — a;" *.
ax~^ +a~^x+ 2
14. Simplify
a*
16. Divide 4 + ^ by 4 + 4
16. Shew that
X
a;i  1 a;8 + 1 a;^  1 aj^ + 1
17. Shew that
(2aj + y') {2y + aj"*) = {2xiyi + ajiy"*)'.
18. Shew that
19. Shew that, if
a;*+2/* + «* = 0, then (x + y + zy27xyz.
20. Find factors which will rationalize the foUowinff
o
expressions :
(i) ai + 6^, (ii) a^x^ + y\
(iii) a + 5ar + caj*, and (iv) x^ + y^ + z.
21. Shew that, if
(la;^)i(y«) + (ly)i (««) + (l»»)i(aj2^) = 0,
and a;, y, z are all unequal, then
CHAPTER XIV.
Surds. Imaginary and Complex Quantities.
171. Definitions. A surd is a root of an arithmetical
number which can only be found approximately.
An algebraical expression such as isja is also often
called a surd, although a may have such a value that ^Ja
is not in reality a surd.
Surds are said to be of the same order when the same
root is required to be taken. Thus V2 and s/Q are called
surds of the second order, or quadratic surds ; also i/4i is a
surd of the third order y or a cubic surd ; and ^a is a surd
of the nth order.
Two surds are said to be similar when they can be
reduced so as to have the same irrational factors. Thus
VB and s/1% are similar surds, for they are equivalent to
2V2 and 3V2 respectively.
The rules for operations with surds follow at once from
the principles established in the previous chapter.
Note. It should be remarked that when a root symbol
is placed before an arithmetical number it denotes only
the arithmetical root, but when the root symbol is placed
before an algebraical expression it denotes one of the roots.
Thus isja has two values but ^2 is only supposed to denote
the arithmetical roofc, unless it is written ± \/2.
214 SURDS.
172. Any rational quantity can be written in the form
of a surd. For example,
2 = ^4 = ^8 = ^2*,
and a = ^a* = ^a' = v^a*.
Also, since tjax. »Jh = \/ab [Art. 165],
we have 2V2 = ^4x^2= V(4 x 2) = ^8,
51^3 = ^5'* X 4/3 = ^{5' X 3) = 4^375,
and a:fab = l/oT x H/ab = v^(a" xab)= ^a^.
Conversely, we have VIS = V(9 x 2) = V9 x V2 = 3V2,
and
^135 + 4/40 = ^(^ X 5) + 4/(2^ X 5) = 3^5 + 2^5 = 5^/5.
173. Any two surds can be reduced to surds of the
same order. For if the surds be ^a and ^/b, we have
j^a = "^ar, and 76 = "76" [Art. 165].
Ex. Which is the greater, 4^14 or 4/6?
The surds most be redaced to equivalent surds of the same order.
Now ^14=4^143=4/196, and 4/6=4/6' =4^216. Hence, as 4^216 is
greater than ,^196, ,^6 must be greater than ,^14.
Thus we can determine which is the greater of two surds without
finding either of them.
174. The product of two surds of the same order can
be written down at once, for we have ^a x JJ/6 = 4^06.
Hence, in order to find the product of any number of
surds, the surds are first reduced to surds of the same
order : their product is then given by the formula
>C/a X yi X 7c...= :;/abc...
Ex. 1. Multiply ^/6 by 4/2.
V5 X 4/2 = 4^53 X 4/22 = 4^(63 X 22) = 4/500.
Ex.2. Multiply 8^6 by 24/2.
3 ^5 X 2^2 = 3 X 2 X ^6 X 4/2= 6 X ^53 X 4^22 = 6 4/500.
MULTIPLtCATTON OF SURDS. 215
fix. 8. Multiply s/2 by f/2.
^/2 X 4^2=^2» X ^2>=/y 2*7^ =4^32.
Or thus: V2x^2=2ix24=2i+i=2*=4/25.
Ex. 4. Multiply ^2+^8 by ^S+^S.
(^/3+^/2)(^/3+^/6)=^/3x^3+V2x^/3+^/3x^/5+^/2x5
=:3+V6+s/15+^/10.
Ex. 6. Divide 4^4 by ^8.
175. The determination of the approximate value of
an expression containing surds is an arithmetical rather
than an algebraical problem ; but an expression containing
surds must always be reduced to the form most suitable
for arithmetical calculation. For this reason when surds
occur in the denominators of fractions, the denominators
must be rationalized. [See Art. 169.]
The foUowing examples will illustrate the process:
^/5.
2__ 2x^/5 ^2
^/5 ^5x^6 6
3 _ 3(^5 + 1) _3
^/5.l"(V5l)(V5 + l)'"4
(n/5 + 1).
176. The product and the quotient of two similar
quadratic surds are both rational.
This is obvious ; for any two similar quadratic surds
can be reduced to the forms a*Jb and c^^b.
Conversely, if the product of the quadratic surds ^/a
and V^ is rational and equal to x, we have x^^is/ax's/b;
therefore xi^b = *^a x ^/b x ^b = b^/a, which shews that the
surds are similar. So also, ii is/a^»Jb is rational, the surds
must be similar.
216 SURDS.
177. The following theorem is important.
Theorem. If aV *Jh^x\ »Jy, where a and x are
rational, and ^/b and ijy are irrational; then will a = fl7,
and b^y.
For we have a — a; + \/6 = «Jy.
Square both sides ; then, s^ter transformation, we have
2 (a — a?) \/6 = y — 6 — (a — ^)'«
Hence, unless the coefficient of /s/b is zero, we must have
an irrational quantity equal ^to a rational one, which is
impossible.
The coefficient of *Jb in the last equation must there
fore be zero, so that a = x. And when a = x, the given
relation shews that V^ = Vy» and therefore b = y.
As a particular case of the above,
Va ^b + fjc, unless 6 = and a = c.
Hence ija + \/c can only be rational when it is zero.
Ex. 1. Shew that »Ja\^b\Je^Ot unless the surds are aU similar.
For we should have J a +/Jb=fJc; and therefore a+b+ 2ijay/b = c.
Hence ^Ja^b is rational, wnieh shews [Art. 176], that ^a and ijb are
similar surds.
178. The expressions a\tJb and a — s/b are said to be
conjugate quadratic surd expressions.
It is clear that the sum and the product of two conju
gate quadratic surd expressions are both rational.
Conversely, if the sum and the product of the expres
sions a\tJb and c\\/d are both rational, then a = c and
^b\tJd = 0, so that the two expressions are conjugate.
For a\c\'s/b\ tsjd can only be rational when *Jb + »Jd
is zero. [Art. 177.]
And, when \/d = '^»Jb, the product (a + V^) (c + V^i)
« oc + (c — a) V& — 6, which cannot be rational unless o = a.
179. In the expression
aoT + 6a?""' + cx""^ + + k,
where a,b, c, k are all rational, let a+ ViS be substi
SURDS. 217
tuted for x; and let P be the sum of all the rational terms
in the result and Q Vi9 the sum of all the irrational terms.
Then the given expression becomes P + Q V^.
Since jr and Q are rational, they contain only squares
and higher even powers of VA ^^^ hence P and Q will not
be changed by changing the sign of \/l3. Therefore when
a — ^/j3 is substituted for a; in the given expression the
resuh will be P  Q V/3.
If now the given expression vanish when a + Vi^ is
substituted for x, we have
Hence, as P and Q are rational and V/3 is irrational,
we must have both P=0 and Q = 0; and therefore
Therefore if the given expression vanish when a + ^J^
is substituted for x it will also vanish when a — »JP is sub
stituted for X,
Hence [Art. 88], if a? — a — V^S be a factor of the given
expression, aj — a + V^ will also be a factor.
Thus, if a rational and integral expression he divisible
by either of two conjugate quadratic surd expressions it
mil also be divisible by the other,
180. The square root of a binomial expression
which is the sum of a rational quantity and a quadratic
surd can sometimes be found in a simple form. The pro
cess is as follows.
To find *J{a + *Jb), where ^Jb is a surd.
Let V(ct + V&) = V^ + Vy*
Square both sides ; then
aits/b^^x + yh 2jxy,
Now, since \/b is a surd, we can [Art. 177] equate the
rational and irrational terms on the different sides of the
last equation; hence x + y = a, and 4^xy = 6.
218 SURDS.
Hence a and y are the roots of the equation
a?* — cw? + 7 = 0,
and these roots are
h {a + V(a*6)} and i {a V(a'6)}.
Thus V(a + Vi) = ^"^^f'^ y "^y^) .
It is clear that, unless \/(a* — 6) is rational, the right
side of the last equation is less suitable for calculation than
the left. Thus the above process fails entirely unless
a* — 6 is a square number; and as this condition will not
often be satisfied, the process has not much practical
utility.
It should be remarked that if os and y are really
rational, they can generally be written down by inspection.
Ex. 1. Find ^(6+2^5).
Let ;^(6 + 2 J5) =s/^+ ijj/. Then, by squaring, we have 6 + 2/^5
= X + y + 2 ^xy' Hence, equating the rational and irrational parts,
x+y = ^ and xy = 5, Whenoe obviouslj x = l and y = 5. Thus
^/(6 + 2^6) = 1 + ^/5.
Ex.2. Find V(28 5^/12).
Let ^{285J12)=Jx'Jy. Then, as before, 4«y=25xl2, or
xy = 75 and a; + y = 28 ; whence x = 25 and y = 8. Thus /^(28  6 v/12)
= 5 ~ ^3. [If we had taken x=S and j^ = 25 we should have had the
negative root, namely ^3  5.]
Ex.3. Find ^/(18 + 12^3).
In this case ^{a^  b) is irrational and therefore the required root
cannot be expressed in the form ^x\Jy where x and y are rational.
The root can however be expressed in the form ^x+i^y; for
V(18 + 12^3)=^/{s/3(12 + 6^/3)}=4/3x^(12 + 6^/3)=4/3x(3+^3)
=4/243 + 4/27.
Ex.4. Find <^(10 + 2^/6 + 2^^10 + 2^16).
Assume V(10+2^6 + 2^10 + 2V16)=^/a;+Jy+J[z;thenl0+2^6
+ 2,^10 + 2<^15 = x + y + z + 2 Jxy + 2ijxz + 2jyz. We have now to
find, if possible t rationtbl values of x, y, z such that xy=^t xz=10,
yzs:15 and x+y+z=10. The first three equations are satisfied by
the values ar=2, y = 3, ;z=6, and these values satisfy x+y\z=10.
Hence ^/(10 + 2^/6 + 2v'10 + 2V15)=^/2 + v'3+V6.
IMAGINARY AND COMPLEX QUANTITIES. 219
Ex. 6. Prove that, if i^(a + \/&) =x+^y; theh will ^{a  ^b) =x »Jy,
We have a+^h=^(x+^yY=ci^+^\^y(Zx^\y),
Hence, eqaating the rational and irrational parts, we have
a=a:8+3a?y, and ^b=^y(3x^+y),
 Hence a^Jb=x^+Bxy^y{Sxz+y);
/. il(a^b)=x^y.
EXAMPLES XVIIL
Simplify the following :
1. 4:^. 2. 2^/5
V3 + 1* s/5 + ^3
5.
V8 + ^3 J^JZ
3^2 4^3 , ^/6
>/3 + V6 ^/6+^2^^2 + ^3•
(72 V5) (5 + ^7) (31 + 13 ^5 )
° (62^7)(3 + V5)(ll + 4V7) •
7 1 8 1
^2 + ^3 + ^6 **• 73+^572
9.
10.
1
^/lO + ^U + ^15 + ^21
1
V6 + V21s/I0^35'
1 1 4 5
4/21"^ 4/2+1* 4'9l"^y9+l'
13. ^ — A — ^ . 14
l + 4,'2+4/4 " ^2 + 4/6+^18*
220 IMAGINARY AND COMPLEX QUANTITIES.
15. V(101  28 V13). 18. V(286V12).
17. V{11 + 2 (1 + V5) (1 + s/7)}.
18. J{6iJ?i+J{lQ8JS)}.
19. V(97.56V3). 20. ^1^^^^.
J2 + J(72J10)'.
22.
23.
^3 + ^2 V^^/2
^2 + ^(2 + V3) ^/2  ^(2 + ^3) *
V(5 + 2^6)^(52^6)
24. 716 + 2^2 + 2^3 + 2^6}.
26. V{11 + 6 V2 + 4 V3 + 2 V6}.
26. V{17 + 4V24V34V64V52V10^2V30}.
27. Shew that
1 1 2
V(12  V140) V(8  V60) "■ V(10 +V84)
28. Show that
1 ^ 4
= 0.
V(l 1  2 V30) V(7  2 VIO) n/(8 + 4 V3)
0.
Imaginary and Complex Quantities.
181. We have already seen that in order that the
formula obtained in Art. 81 for the factors of a quadratic
expression may be applicable to all cases, it is necessary
to consider expressions of the form J —a, where a is
IMAGINARY AND COMPLEX QUANTITIES. 221
positive, and to assume that such expressions obey all the
fundamental la^vs of algebra.
Since all squares, whether of positive or of negative
quantities, are positive, it follows that J— a cannot
represent any positive or negative quantity; it is on this
account called an imaginary quantity. Also expressions
of the form a + 6 v — 1 where a and b are real, are called
complex quantities.
182. The question now arises whether the meanings
of the symbols of algebra can be so extended as to include
these imaginary quantities. It is clear that nothing would
be gained, and that very much would be lost, by extending
the meaninga of the symbols, except it be possible to do
this consistently with all the fundamental laws remaining
true.
Now we have not to determine all the possible systems
of meanings which might be assigned to algebraical
symbols, both to the symbols which have hitherto been
regarded as symbols of quantity and to the symbols of
operation, subject only to the restriction that the funda
mental laws should be satisfied in appearance whatever the
symbols may mean: our problem is the much simpler and
more definite one of finding a meaning for the imaginary
expression >/— a which is consistent with the truth of all
the fundamental laws.
183. We already know that — 1 is an operation which
performed upon any quantity changes it into a magnitude
o f a d iametrically opposite kind. And, if we suppose that
V— 1 obeys the law expressed by 1 x J— 1 x >/^^= — 1,
it follows that v — 1 must be an operation which when
repeated is equivalent to a reversal.
Now any species of magnitude whatever can be re
presented by lengths set off along a straight line; and,
when a magnitude is so represented, we may consider the
222 COMPLEX QUANTITIES.
operation J— 1 to be a revolution through a right angle,
for a repetition of the process will turn the line in the
same direction through a second right angle, and the line
will then be directly opposite to its original direction.
Hence, when magnitudes are represented by lengths
measured along a straight line, we see that >/—l, regarded
as a symbol of operation, has a perfectly definite meaning.
The symbol >/— 1 is generally for shortness denoted by
i, and the operation denoted by i is considered to be a
revolution through a right angle counterclockwise, — %
denoting revolution through a right angle in the opposite
direction.
184. It is clear that to take a units of length and
then rotate through a right angle counterclockwise gives
the same result as to rotate the unit through a right angle
counterclockwise and then multiply by a. Thus ai = ia.
Again, to multiply ai by bi is to do to ai what is done
to the unit to obtain bi, that is to say we must multiply
by 6 and then rotate through a right angle; we thus
obtain ab units rotated through two right angles, so that
ai xbi=' ab = ahii.
From the above we see that the symbol i is commuta
tive with other symbols in a product.
Since (ai) x {ai) = aaii = a^ ( 1) = — a^ it follows that
J^a*==ai; it is therefore only necessary to use one
imaginary expression, namely «s/— 1.
185. With the above definition o{ J1 ori, namely
that it represents the operation of turning through a right
angle counterclockwise, magnitudes being represented by
lengths measured along a straight line, the truth of the
fundamental laws of algebra for imaginary and complex
expressions can be proved. Some simple cases have been
considered in the previous Article: for a full discussion
see De Morgan's Double Algebra; see also Clifford's
Common Sense of the Exact Sciences, Chapter iv. §§ 12
and 13.
CONJUGATE COMPLEX EXPRESSIONS. 223
186. If a + bi = 0, where a and b are real, we have
a = — bi. But a real quantity cannot be equal to an im
aginary one, unless they are both zero.
Hence, if a + bi = 0, we have both a = and 6 = 0.
Note. In future, when an expression is written in
the form a + bi, it will always be understood that a and 6
are both real.
187. If a + 6i = c + di, we have a — c + (6 — d) i = ;
and hence, from Art. 186, a — c = and 6 — d = 0.
Thus, two complex expressions cannot be eqttal to one
another, unless the real and imaginary parts are separately
equal,
188. The expressions a 4 bi and a — bi are said to be
conjugate complex expressions.
The sum of the two conjugate complex expressions
a^bi and a — 6i is a + a + (6 — fe)i=2a; also their pro
duct is oa + dbi — abi — 6V = a^\ b\
Hence the sum and the product of two conjugate complex
expressions are both real.
Conversely, if the sum and the product of two complex
expressions are both real, the expressions must be con
jugate.
For let the expressions be x + bi and c + di. The sum
iaa + bi + c + di = a + c+{b + d)i, which cannot be real
unless b + d = 0. Again,
(a + bi) (c f di) = ac+bci+adi + 6c?i* = ac — bd'\ (be h ad)i,
which cannot be real unless 6c + od = 0. Now, if 6 + d =
and also bc + ad = 0, we have b (c — a) = ; whence
a = c or 6 = 0. If 6 = 0, d! is also zero, and both expres
sions are real ; and, if 6 + 0, we have a=^c, which with
6 = — d, shews that the expressions are conjugate.
189. Definition. The positive value of the square
root of a' + 6' is called the modulus of the complex
224 MODULUS OP A COMPLEX EXPRESSION.
quantity a + U, and is written mod (a + hi). Thus
mod (a + W) = + Ja^ + i*.
It is clear that two conjugate complex expressions have
the same modulus ; also, since (a + hi) (a — 6i) = a* + 6*
[Art. 188], the modulus of either of two conjugate complex
expressions is equal to the positive square root of their
product.
Since a and h are both real, a' + 6' will be zero if, and
cannot be zero unless, a and^ 6 are both zero. Thus the
modulus of a complex expression vanishes if the expression
vanishes, and conversely the expression will vanish if the
modulus vanishes.
If in mod (a + hi) = + «/a* + 6' we put 6 = 0, we have
mod a = + s/a", so that the modulus of a real quantity is
its absolute value.
190. The product of a + 6i and c + di is
aC'\hci'\adi\hdi^=^aC'hd\(bc\'ad) i.
Hence the modulus of the product of a + hi and
c + diia
ij{{ac  hd)' + {he + ad)'} = V{(a' + 6*) (c« + (P)}
Thus the modulus of the product of two complex
expressions is equal to the product of their moduli
The proposition can easily be extended to the case of
the product of more than two complex expressions ; and,
since the modulus of a real quantity is its absolute value,
we have the following
Theorem. The modulus of the product of any nwmber
of quantities whether real or complex, is equal to the
product of their moduli.
191. Since the modulus of the product of two com
plex expressions is equal to the product of their moduli, it
follows conversely that the modulus of the quotient of two
expressions is the quotient of their moduli. This may
also be proved directly as follows :
MODULUS OF A PRODUCT. 225
. . , .V , 7 .V a\hi c — di
_ac + bd + (bc — ad)i
Hence mod ^ = VK^ + ^^)' + (^^ " ^'}
[c + di) c +ar
_ V{a^ + y} _ inod(a + 60
" V{c* + d'i " mod (c + diy
192. It is obvious that in order that the product of
any number of real factors may vanish^ it is necessary
and sufficient that one of the factors should be zero, and,
by means of the theorem of Art. 190, the proposition can
be proved to be true when all or any of the factors are
complex quantities.
For, since the modulus of a product of any number of
factors is equal to the product of their moduli, and since
the moduli are all real, it follows that the modulus of
a product cannot vanish unless the modulus of one of its
factors vanishes.
Now if the product of any number of factors vanishes
its modulus must vanish [Art. 189] ; therefore the modu
lus of one of the factors must vanish, and therefore that
factor must itself vanish. Conversely, if one of the
factors vanishes, its modulus will vanish; and therefore
the modulus of the product and hence the product itself
must vanish.
193. In the expression
where a, b,Cy,,,k are all real, let a + ^i be substituted for as,
and let P be the sum of all real terms in the result, and
Qi the sum of all the imaginary terms. Then the given
expression becomes P + Qi.
Since P and Q are both real, they can contain only
s. A, 16
226 MODULUS OF A PRODUCT."
squares and higher even powers of i, and hence P and Q
will not be changed by changing the sign of i Therefore
when a — ^i is substituted for x in the given expression
the result will be P — Qi.
If now the given expression vanishes when a + pi is
substituted for a?, we have P + Qi = 0.
Hence, as P aod Q are real, we must have both P =
and Q = 0, and therefore P — Qi = 0.
Hence if the given expression vanishes when a + /8i
is substituted for x, it will also vanish when a — /8i is
substituted for x.
Therefore [Art. 88] if x — a — jSi is a factor of the
given expression, x — a + lSi will also be a factor.
Thus, if any expression rational and integral in x,
and with all its coefficients real, be divisible by either of
two conjugate complex expressions it will also be divisible
by the other.
r
i
CHAPTER XV.
Square and Cube Roots.
194. We have already shewn how to find the square
of a given algebraical expression; and we have now to
shew how to perform the inverse operation, namely that
of finding an expression whose square will be identically
equal to a given algebraical expression. It will be seen
that our knowledge of the mode of formation of squares
will enable us in many cases to write down by inspection
the square root of a given expression.
195. From the identity
a'±2ab + F = (a±b)\
we see that when a trinomial expression consists of the
sum of the squares of any two quantities plus (or minus)
twice their product, it is equal to the square of their sum
(or difference).
Hence, to write down the square root of a trinomial ex
pression which is a perfect square, arrange the expression
according to descending powers of some letter ; the square
root of the whole expression will then be found by taking
the square roots of the extreme terms with the same or
with different signs according as the sign of the middle
term is positive or negative. •
Thus, to find the square root of
4a«  12a*6» + 9b\
15—2
228 SQUARE ROOT.
The square roots of the extreme terms are ± 2a* and
± 36'. Hence, the middle term being negative, the re
quired square root is ± (2a* — 36').
Note. In future only one of the two square roots of
an expression will be given, namely that one for which
the sign of the first term is positive : to find the other
root all the signs must be changed
196. When an expression which contains only two
diflferent powers of a particular letter is arranged accord
ing to ascending or descending powers of that letter, it will
only consist of three terms. For example, the expression
a^ + 6' + c* + 26c + 2ca + 2a6 when arianged according to
powers of a is the trinomial
a' + 2a (6 + c) + (6» + c' + 26c).
It follows therefore from the preceding article that
however many terms there may be in an expression which
is a perfect square, the square root can be written down
by inspection, provided that the expression contains only
two different powers of some particular letter.
Ex. 1. To find the sqnare root of
a'+ 6'+ c2+ 26c + 2ca + 2a6.
Arranged according to powers of a, we have
a2+2a(6 + c) + {6 + c)2, thatis {a+(6+c)}«.
Hence the required square root is a + & + c
Ex. 2. To find the square root of
4a:< + 9y* + 16^ + 12xhi^  l^xh^  24:yh\
The given expression is
4a:* + 4a;« (% «  4«3) + 9y4 _ 24y*22 + 162!*,
that is, (2x2)8 + 2 (2x2) (%3  48«) + (V  4«2)2,
which is {2x«+(3y«4e2)}2.
Hence the required square root is 2x2+3^*  ^\
SQUARE BOOT. 229
Ex. 3. To find the square root of
a2+2a6a; + (6a+2ac)a;a+26c«*+c«ic*.
Arrange according to powers of a; we then have
a^+2a{bx + cx^) + If^x^ + 2hcix^ + c^x*,
that is, a^ + 2a (6x + ca?) + (bx + cx^)K
Hence the required square root is a + bx + cz\
Ex. 4. To find the square root of
a;6_2a^ + 3a;4+2a:«(yl) + x2(l2y)+2a:y+ya.
The expression only contains y^ and y ; we therefore arrange it
according to powers of y^ and have
y«+2y(a;3_a.2+g(.)^.a.6_2a:C + 3p4_2x»+x«.
Now, if the expression is a complete square at all, the last of the
three terms must be the square of half the coefficient of y ; and it is
easy to verify that
(afiQ^+xf=afi2at^+^xl^'2a?¥xK
Hence the required square root is y\'a^^\x»
197. To find the square root of any algebraical ex
pression.
Suppose that we have to find the square root of (4 + Bf,
where A stands for any number of terms of the root, and B
for the rest; the terms mA and B being arranged accord
ing to descending (or ascending) powers of some letter, so
that every term in A is of higher (or lower) degree in that
letter than any term of B,
Also suppose that the terms in A are known, and that
we have to find the terms in B,
Subtracting A^ from (A + J?)', we have the remainder
(%A+B)B.
Now from the mode of arrangement it follows that the
term of the highest (or lowest) degree in the remaiuder is
twice the product of the first term in A and the first term
in B,
Hence, to obtain the next term of the required root, that
230 SQUARE ROOT.
is, to obtain the highest (or lowest) term of B, we subtract
from the whole expression the square of that part of the root
which is already found, amd divide tiie highest (or lowest)
term of the remainder by twice the first term ofihe root
The first term of the root is clearly the square root of
the first term of the given expression; and, when we have
found the first term of the root, the second and other terms
of the root can be obtained in succession by the above
process.
For example, to find the square root of
The process is written as foUows :
We first take the square root of the first term of the given
expression, which must he arranged according to ascending or <2e
seending powers of some letter: we thus obtain sfi, the first term of
the required root.
Now subtract the square of sfi from the given expression, and
divide the first term of the remainder, namdy 4a^, by 2a^: we
thus obtain  2^, the second term of the root.
Now subtract the square of a^2^ from the given expression,
and divide the first term of the remainder, namely 2x^, hyia^i we
thus obtain z, the third term of the root.
Now subtract the square of x?  2x*+x from the given expression,
and divide the first term of the remainder, namely 4i^, by 2;r': we
thus obtain  2, the fourth term of the root.
Subtract the square of 3fi'23^+x2 from the given expression
and there is no remainder.
Hence s^^a^+x 2 is the required square root.
The squares of re*, afi'2x\ <fec. are placed under the given
expression, like terms being placed in the same column, so that in
every case the first term of the remainder is obvious.
«
198. The square root of an algebraical expression may
also be obtained by means of the theorem of Art 91.
Take for example the case just considered.
SQUARE ROOT. 231
The required root will be aa? ^ba? •\cx\dy provided
that the given expression is equal to (cw?' + 6a?* + ca? + d)*,
that is equal to
aV + 2a6aj* + (2ac + 6*) ^* + 2 (ad + 6c) aj*
+ (26d + c*) aj^ + 2cda7 + cT.
, Hence, equating the coeflScients of corresponding
powers of x in the last expression and in the expression
whose root is required, we have
a"=l; 2a6 = 4; 2ac + 6*=6; 2ad! + 26c = 8;
26d+c' = 9; 2cd = 4; (? = 4.
The first four of these equations are sufficient to
determine the values of a, h, c, d; these values are (taking
only the positive value of a), a = 1, 6 = — 2, c = 1, d = — 2.
The last three equations will be satisfied by the values
of a, 6, c, d found from the first four, provided the given
expression is a perfect square, which is really the case.
Thus the required square root is x^ — 2x^ + a? — 2.
199. Extended Definition of Square Root. The
definition of the Square Root of an algebraical expression
may be extended so as to include the case of an expression
which is not a perfect square: For, although an expres
sion may not be a perfect square, we can find, by the
methods of Art. 197 or Art. 198, a second expression
whose square is equal to the given expression so far as
certain terms are ^ncemei * ^
Thus the square root of a?'\^x may be said to be
flj + 1, (a? + 1)' being equal to a?* + 2a? so far as the terms
which contain x are concerned.
Again, the square root of 1 + a? may be said to be
l + ^orl + Q— Q, the square of the former differing from
X^
l + x by T > ^^^ t,he square of the latter differing by
232 SQUABE BOOT.
— \a? + ■^^x*'. Thus, provided x is small, 1 + 5 is an
approximation to the square root of 1 + ^, and 1 + 5 — q
is a closer approximation, and by continuing the process
we can approximate as closely as we please to the square
root of 1+x; this however is by no means the case when
X is not a small quantity.
200. When any number of terms of a square root have
been obtained as many more can be found by (yrdinary
division.
For suppose the expression whose square root is to be
found is the square of
The coefficients a^,a^,,.,a^ can be found by equating
the coefficients of the first 2r powers of x in the square of
the above to the coefficients of the corresponding powers
of X in the given expression.
The square of the above expression is
+ 2 JJ (a^+,a?""" + . . . + a^x^^"^') + iJ*].
Now, since the highest power of a? in JS is a?"*^, the
highest power of x in the expression within square brackets
Hence the expression within square brackets will not
affect any of the terms from which a^, a^, ..,a^ are deter
mined, for the first 2r terms of the given expression ex
tend from a?"" to a?^^\
It therefore follows that if the square of the sum of
the first r terms of the root be subtracted from the given
SQUARE ROOT. 233
expression, and the remainder be divided by twice the
sum of the first r terms, the quotient will give the next
r terms of the root.
201. When n figures of a square root of a , number
have heenfoimd by the ordinary method, n— 1 m^yre figures
can be fotmd by division, provided that the nwmber is a
perfect square o/ 2w — 1 figures ; if however this be not the
case, there may be an error in the last figure.
Let N be the given number, which is the perfect
square of a number containing 2n — 1 figures, and let p
be the number formed by the first n figures followed by
n — 1 zeros, and let q be the number formed by the
remaining w — 1 figures.
Then a/N =jp + g ;
.'. (Np')/2p=^q + ^/2p.
Now 2p <t 2 . 10*~"' and g > 10**''. Hence qy2p must be a
fraction ; whence it follows that if p^ be subtracted from
N and the remainder be divided by 2p, the integral part
of the quotient will be q.
Next, let ^N contain m figures, where m is greater
than 2w — 1.
Let p be the number formed by the first n figures of
the root followed by m — n zeros, let q be the number
formed by the next w — 1 figures followed by m — 2w + 1
zeros, and let r be the number formed by the m — 2n + 1
remaining figures. Then
N={p{q + ry;
.'. (N  p^)l2p  y = (2* + r^ + 2qr)/2p + r.
Now 10"*>p<t;10"*"S
lO*""" > 2 < 10"'"""\
and 10'"'"^^>r<tlO'"^",
234 CUBE ROOT.
whence it follows that (g^ + r* + 2 jr)/2p is less than
Hence (g' + r* + 2gr)/2p + r is less than 2 x 10*^**,
but it is not necessarily less than 10"*'"*'*"\ Hence
(iV— j5')/2p may dififer from q by more than 10"*"***^; it
must however dififer by less than 2 x 10**"^*^; so that the
n — 1 first figures of the quotient (iV—)*)/2p are either
the 71—1 figures of q or dififer only in the last figure,
and in that case by 1 in excess.
Cube Root.
202. From the identity
(a + 6)' = a' + 3a'6 + 3a6' + h\
we see that the cube of a binomial expression has fcmr
terms, and that when the cube is arranged according to
ascending or descending powers of some letter, the cube
roots of its extreme terms are the terms of the original
binomial.
Hence the cube root of any perfect cube which has
only four terms can be written down by inspection, for we
have only to arrange the expression according to powers
of some letter and then take the cube roots of its extreme
terms.
For example, if 27a'^64a'6+36a*628a*6* is a perfect cnbe its
cabe root must be 3a^  2db ; and by forming the cube of 3a^  2ab
it is seen that the given expression is really a perfect cube.
When an expression which contains only three diflferent
powers of a particular letter is arranged according to
powers of that letter, there will be only^Mr terms.
It therefore follows that however many terms there
may be in an expression which is a perfect cube, the cube
root can be written down by inspection, provided that the
CUBE ROOT. 235
ft
expression contains only three different powers of some
particular letter.
For example, to find the cube root of
a' + 6» + c8 + 3a«6 + 3a«c + 3a62 + 3ac2 + 6a6c + SftSc + 36c».
Arranged according to powers of a, we have
o»+3a3(6 + c) + 3a(6a+ca+26c) + 6»+c» + 362c + 36c>,
that is, a»+3a>(6+c) + 3a(6+c)2+(6+c)S.
Hence the required root is a+b\c.
203. To find the cicbe root of any algebraical expression.
Suppose we have to find the cube root of {A \ By,
where A stands for any number of terms of the root, and
B for the rest; the terms in A and B being arranged
according to descending (or ascending) powers of some
letter, so that every term of A is of higher (or lower)
degree in that letter than any term of B.
Also suppose the terms in A are known, and that we
have to find the terms in B.
Subtracting A^ firom (A H By, we have the remainder
(3A' + SAB'^B')B.
Now from the mode of aiTangement it follows that the
term of the highest (or lowest) degree in the remainder is
3 X square of the first term of 4 x first term of B.
Hence to obtain the neoct term of the required root,
that is, to obtain the highest (or lowest) term of B we
subtract from the whole eaypression the cube of that part
of the root which is already found and divide the highest
(or lowest) term of the remainder by three times the square
of the first term of the root.
This gives a method of finding the successive terms
of the root after the first ; and the first term of the root
is clearly the cube root of the first term of the given
expression.
236 CUBE ROOT.
For example, to find the cabe root of
ofi  6a:»y + 21af»ya  44a;8y» + 63ar»y*  64xy» + 27y«.
The process is written as follows :
a:«  6x5?/ + 21a:^*  44xV + 63a; V  54«y» + 27y«
(a?  2a;y )» = a :«  6x8y f 12g^»  8a:^»
(a;2  2a:y + 3y 2)3 = a:6 _ 6a*y + 21a*y 2  44a:32^» + 63a%*  64«y * + 2
Having arranged the given expression according to descending
powers of x, we take the cabe root of the first term: we thus obtain
a;*, the/r«< term of the required root.
We then subtract the cabe of oe^ from the given expression, and
divide the first term of the remainder, namely  BaH'y, by 3 x {t^)^ :
we thus obtain  'Ixy^ the second term of the root.
We then subtract the cube of a^  2xy from the given expression,
and divide the first term of the remainder by 3 x {x*)* : this will give
tiie third term of the root.
Note. The above rule for finding the cube root of an
algebraical expression is rarely, if ever, necessary.
In actual practice cube roots are found as follows.
Take the case just considered ; the first and last terms
of the root are a^ and St/*, the cube roots of the first and
last terms of the given expression ; also the second term
of the root will be found by dividing the second term of
the given expression by 3 x {x'^Y, so that the second term
of the root is — 2ay.
Hence, if the given eocpression is really a perfect cube,
it must be {a? — £cy + 3y% and it is easy to verify that
{x' — 2a?y + ^yy is equal to the given expression.
Again, to find the cube root of
of  Qx'^y + 15a?y  29^y h Sla^y  60a?y + 64a;y
 63^y + 27ay  27y».
If the given expression is really a perfect cu be the
first and last terms of the root must be ^a? and v^— 27t/'
respectively, that is a^ and — Sy*.
EXAMPLES 237
The second term of the root must be — Qx^y — 3 {a?Y
= — ^y ; and the term next to the last must be
27a;2/' T 3 ( 3^)' = + ajy^
Hence the given expression, if a cube at ally
must be {a? — 2x^y H xy^ — 3^")' ; and by expanding
{a? — 2a?y + scy* — 33/^)® it will be found that the given
expression is really a perfect cube.
204. From the identity [see Art. 253]
(a + hy = a" + na'^'^b + terms of lower degree in a,
it is easy to shew, as in Articles 197 and 203, that, the
w*^ root of any algebraical expression can be found by
the following
Rule. Arrange the expression according to descending
or ascending powers of some letter, and take the n*^ root of
Hie first term : this gives the first term of the root.
AlsOy having found any number of terms of live root,
subtract from the given expression the w*^ power of that
part of the root which is already found, and divide the first
term of the remainder by n times the (n — 1)*^ power of the
first term of the root : this gives the next term of the root
EXAMPLES XIX.
Write down the square roots of the following expressions :
1. 4a;^*»12aj«y + 9y^
2. «f» + 9a;V"  6a;y .
3. a* + 46' + 9c' + 1 2hc  6ca  iah.
4. 25a* + 96* + 4c* + 1 26V  20c V  30a«6'.
238 EXAMPLES.
Find the square roots of
6. 4aj*8a:'y* + 4£cy' + y".
7. 49 + 1 1 2a;» + IOt? + 64a;* + SOa;* + 25a:^
8. a;*2a;» + 5a;'6a; + 86a;» + 5aj"2a;* + a;*.
9. ?^^_^ 2054^ + 2.
y* 25ar y ox
10. flc*  4a;^ + 2aj + 4a5^ + x^.
11. a;*  4a;' + 4a; + 2^^  4a;^ + a;^
12. a;*  2a;"'^a;^ + 2a^a;* + a~^x^  2a^a;^ + a\
Find the cube roots of
13. a;'24a;'+192a;512.
14. a;*  3a;V + 6a;y  7a;'2/« + 6a;y  Say* + y\
15. 1  9a;* + 33a;*  63a;« + 66a;»  36a;^» + 8a;'^
16. Find the square root of
2a»(6 + c)' + 26»(c + a)» + 2c'(a + 6)» + 4a6c(a + 6 + c).
17. Find the square root of
a;* (a;* + y* + «*) + y^s? + 2a; (y + z) {yz  a;).
18. Find the square root of
(a6)*2(a« + 6'')(a6)' + 2(a* + 6*).
19. Shew that (a; + a) (a; + 2a) (a; + 3a) (a; + 4a) + a* is a
perfect square.
20. Prove that a;* + px^ + qx^ + rx + 8 is a square, if p^a = r*
aiidf>'4pg' + 8r = 0.
EXAMPLES. 239
21. Find the values oi A^ B and C in order that
may be a perfect square.
22. Shew that, if cuxf + bQi:^ + cx + d be a perfect cube, then
6* = Zac and c' = Zbd.
23. Find the conditions that
aa^ + bj^ + cz' + 2/yz + 2gzx + 2hxy
may be the square of an expression which is rational in x, y
and z.
24. Shew that if
(a  X) ic* + (6  X) 2^^ + (c  X) 2' + 2fyz + 2gzx + 2kxy
be the square of an expression which is rational in a;, ^ and z,
then will
a_^^ = 6.¥=e: = X.
J 9 f^
25. Shew that when the first r terms of the cube root of
an algebraical expression are known, r more terms can be
found by ordinary division.
26. When n+2 figures of the cube root of a number have
been obtained by the ordinary method, n more can be obtained
by ordinary division, provided the number is a perfect cube of
27t + 2 figures.
27. Shew that, if n + 2 figures whose numerical value
is a have been found of a positive root of the equation
a? + qxr = Of q being supposed positive, then the result of
dividing rqaa^ by da^ + q will give at least n—l more
figures correctly.
/^
CHAPTER XVL
Ratio. Proportion.
205. Definitions. The relative magnitude of two
quantities, measured by the number of times the one
contains the other, is called their ratio.
Concrete quantities of different kinds can have no
ratio to one another: we cannot, for example, compare
with respect to magnitude miles and tons, or shillings
and weeks.
The ratio of a to 6 is expressed by the notation a \h\
and a is called the first term, and h the second term, of the
ratio. Sometimes the first and second terms of a ratio are
called respectively the antecedent and the consequent.
It is clear that a ratio is greater, equal or less than
unity according as its first term is greater, equal or less
than the second. A ratio which is greater than unity is
sometimes called a ratio of greater inequality, and a ratio
which is less than unity is similarly called a ratio of less
inequality.
The ratio of the product of the first terms of any
number of ratios to the product of their second terms, is
called the ratio compounded of the given ratios.
Thus ac : hd is the ratio compounded of the two ratios a : 6 and c : d.
The ratio a' : 6* is sometimes called the duplicate ratio
of a : 6 ; so also a' : 6', and s/a : s/h are called respectively
the triplicate, and the svh duplicate ratio of a : 6.
BATIO. 241
206. Magnitudes must always be expressed by means
of numberSy and the number of times which one number
contains another is found by dividing the one by the other.
Thus ratios can be expressed 2i& fractions.
The principal properties of fractions and therefore of
ratios have already been considered in Chapter vra.
Thus, a ratio is unaltered in value by multiplying each
of its terms by the same number. [Art. 107.]
Different ratios can be compared by reducing to a
common denominator the fractions which express their
values. [Art. 109.]
The theorems of Art. 113 are also true for ratios.
The following theorem is of importance :
207. Theorem. Any ratio is made more nearly
eqiml to unity by adding the same positive quantity to each
of its terms.
By adding x to each term of the ratio a : 6, the ratio
a + a? : 6 + a? is obtained.
Now T — 1 = 1 — , and j— 1 = j— — ,
6 b b\x b + x
7
and it is clear that the absolute value of y is less than
b + x
that of — ^— , for the numerators are the same and the
b
denominator of the former is the larger: this proves the
proposition.
When X is very great, the fraction , is very small ;
and f , which is the difference between ,— — and 1,
b + x' b + x
can be made less than any assignable diffei'ence by taking
X suflBciently great.
s. A. 16
242 RATIO.
This is expressed by saying that the limiUng value of
T , when X is infinite, is unity.
Now two quantities, whether finite or not, are eqiuil
to one another when their ratio is unity. Thus a + ^ and
b\x are equal to one another when x is infinite, a being
supposed not equal to b. [See Art. 118.]
208. Since any ratio is made more nearly equal to
unity by the .addition of the same quantity to ea<5h of its
terms, it follows that a ratio is diminished or increased by
such addition according as it was originally greater or less
than unity. This proposition is sometimes enunciated :
A ratio of greater inequality is diminished and a ratio of
less inequality is increased by the addition of the same
quantity to each of its terms.
209. Incommensurable numbers. The ratio of
two quantities cannot always be expressed by the ratio of
two whole numbers ; for example, the ratio of a diagonal
to a side of a square cannot be so expressed, for this ratio
is aJ2 : 1, and we cannot find any fraction which is exactly
equal to aJ2.
Magnitudes whose ratio cannot be exactly expressed
by the ratio of two whole numbers, are said to be in
comm,ensurable.
Although the ratio of two incommensurable numbers
cannot be found exactly, the ratio can be found to any
degree of approximation which may be desired ; and the
different theorems which have been proved with respect
to ratios can, by the method of Art. 163, be proved to be
true for the ratios of incommensurable numbers.
proportion. 243
Proportion.
210. Four quantities are said to be proportional when
the ratio of the first to the second is equal to the ratio of
the third to the fourth.
Thus a, b, c, d are proportional, if
a : b = c : d.
This is sometimes expressed by the notation
a :b :: c : d,
which is read " a is to 6 as c is to d"
The first and fourth of four quantities in proportion,
are sometimes called the eoctremes, and the second and
third of the quantities are called the means.
211. If the four quantities a, 6, c, d are proportional,
we have by definition,
a ^c
I'd'
Multiply each of these equals by bd ; then
ad = bc.
Thus the product of the extremes is equal to the product
of the means.
Conversely, if ad = be, then a, 6, c, d will be propor
tional.
For, if
ad =
= be, then
ad
bd'
bo
''bd'
a
•*• b''
c
"5*
a :b'
= c :d.
that is
16—2
244 PROPORTION.
Hence also, the four relations
a :b = c : d,
a : c = b : d,
b : a = d : Cf
and b : d = a : Cf
are all true, provided that ad = be. Hence the four
proportions are all true when any one of them is true.
Ex. If a : 6=c : d, then will a+b : ab^c + d : cd.
This has already heen proved in Art. 113 : it may also be proved
as follows:
a + b : ab=c + d: cdt
if (a + 6)(cd) = (a6)(c + d),
that is, {/ acbd+bcad=ac'bdbc+adi
or, if bc=:ad.
Bat be is eqaal to ad^ since a : b=iC : d.
212. Quantities are said to be in continued propoHion
when the ratios of the first to the second, of the second
to the third, of the third to the fourth, &c., are all equal.
Thus a, b, c, d, &c. are in continued proportion if
a:b=:b : c=^ c : d = &a,
that is, if .= =  = &c.
bed
If a : 6 = 6 : c, then b is called the mean proportional
between a and c; also c is called the third proportional
to a and 6.
I{a,h,c be in continued proportion, we have
a _b
.'. V = a;C, or 6 = ^oc.
PROPORTION. 245
Thus the mean proportional between two given quantities
is the square root of their product,
. , a h h h
Also r X  =  X  »
c c
a 6* a'
VllOlV IS u T 9 •
C C
Thus, if ^A?'^e quantities are in continued propoHion,
the ratio of the first to the third is the duplicate ratio of the
first to the second,
213. The definition of proportion given in Euclid is
as follows: Four quantities are proportionals, when if any
equimultiples whatever be taken of the first and the third,
and also any equimultiples whatever of the second and
the fourth, the multiple of the third is always greater
than, equsd to or less than the multiple of the fourth,
according as the multiple of the first is greater than, equal
to or less than the multiple of the second.
If the four quantities a, b, c, d satisfy the algebraical
a G
test of proportionality, we have t = ;^ ; therefore for all
, « J ma mc
values 01 m and w, — ? = "t*
no na
> . >
Hence mc = wd, according as ma = nb. Thus a, 5, c, d
< <
satisfy also Euclid's test of proportionality.
Next, suppose that a, b, c, d satisfy Euclid's definition
of proportion.
If a and b are commensurable, so that a: b = m:n,
where m and n are whole numbers ; then
am y
T = — ; .'. na — mb,
n
246 PROPORTION.
But by definition
> . >
nc = md according as na = mK
< <
Hence nc = 7nd\
c _^_a
* ' d n b'
Thus a, h, c, d satisfy the algebraical definition.
If a and b are incommensurable we cannot find two
whole numbers m and n such that a: b=^ m: n. But, if
we take any multiple na of a, this must lie between two
consecutive multiples, say mb and (m H 1) 6 of 6, so that
na > mb and na<(m{l)b.
Hence by the definition,
nomd and 7ic<(m + l)d.
Hence both r and ^ lie between — and .
o a n n
Thus the difference between t and % is less than  ; and
b d n
as this is the case however great n may be, ^ must be equal
to T , for their difference can be made less than any
assignable difference by suflSciently increasing n,
Ex. 1. For what value of x wiU the ratio 7+a; : 12 + » be eqaal to
the ratio 5 : 6? Am, 18.
Ex. 2. If 6a;2+%*=13a;y, what is the ratio of a; to y?
Ana, 2 : 3 or 3 : 2.
Ex. 3. What is the least integer which when added to both terms of
the ratio 5 : 9 wiU make a ratio greater than 7 : 10? Ans. &,
Ex. 4. Find x in order that x + l : a; + 6 may be the dupUcate ratio of
29
8 : 5. Ans* r^ .
10
VARIATION. 247
Ex. 5. Shew that, ifa:b::c:d, then
(i) a^+ab + b^: c^+cd + d^:: aPab + b^ : c^cd+d?.
(u) a+b:c + d:: ^(2a«  362) . ^(2c«  Sd^).
(iii) a«+62+c2+(P : {a+b)^+{c + d)^ :: (a+c)«+(6 + d)«
: (a + 6 + c + d)'.
[See Art. 113.]
Ex. 6. Ji a : b :: c : d, then will ab{cd be a mean proportional
between a' + c* and d'* + d*.
Variation.
214. One magnitude is said to vary as another when
the two are so related that the ratio of any two values of
the one is equal to the ratio of the corresponding values
of the other.
Thus, if a,, a, be any two measures of one of the
quantities, and 6^, 6^ be the corresponding measures of the
other, we have
~ = i^ : and therefore r^ = y^.
Hence the measures of corresponding values of the two
magnitudes are in a constant ratio.
The symbol « is used for the words varies as: thus
AccB is read *A varies a& B\
If a oc 6, the ratio a : 6 is constant ; and if Ave put m
for this constant ratio, we have
^ = m; .*. a — Tvih,
b
To find the constant m in any case it is only necessary
to know one set of corresponding values of a and 6.
a 15
For example, if a x &, and a is 15 when b is 5, we have  = 7»=— ;
' .% a =36.
248 VARIATION.
215. Definitions. One quantity is said to vary in
versely as another when the first varies as the reciprocal
of the second.
Thus a varies inversely as 6 if the ratio a : r is constant,
and therefore a6 = m.
One quantity is said to vary as two others jointly when
the first varies as the product of the other two. Thus a
varies as h and c jointly if a oc 6c, that is if a = mbc,
where m is a constant.
One quantity is said to vary directly as a second and
inversely cw a third when the ratio of the first to the
product of the second and the reciprocal of the third is
constant.
Thus a is said to vary directly as^ and inversely as c,
if a : 6 X ^ is constant, that is, if a = m , where m is a
c
constant.
In all the different cases of variation defined above,
the constant will be determined when any one set of
corresponding values is given.
For example, if a varies jointly as h and c ; and if a is 6 when b
is 4 and c is 8, we have
and \6=mx4x3.
Hence m=^ , and therefore a=» 6c.
216. Theorem. If a depends only on b and c, and
if a varies as b when c is constant, and varies as c when
is constant; then, when both b and c vary, a will vary
as be.
Let a, b, c; a\ b', c and a'\ b', d be three sets of
corresponding values.
VARTATION. 249
Then, since c is the same in the first and second, we
^^^« f' = F ^^
And, since V is the same in the second and third, we
have T/ = 7 (ii).
a c
Hence from (i) and (ii), — , = rr; ,
^ ^ a be
which proves the proposition.
The following are examples of the above proposition.
The cost [C] of a quantity of meat varies as the price [P] per
pound if the weight [W] is constant, and the cost varies as the
weight if the price per pound is. constant. Hence, when both the
weight and the price per pound change, the cost varies as the
product of the weight and the price.
Thus, if (7 a P, when W is constant,
and C ccW, when P is constant;
then C a PW, when both P and W change.
Again, the area of a triangle varies as the base when the height
is constant; the area also varies as the height when the base is
constant; hence, when both the height and the base change, the
area will vary as the base and height jointly.
Again, the pressure of a gas varies as the density when the
temperature is constant; the pressure also varies as the absolute
temperature when the density is constant ; hence when both density
and temperature change, the pressure will vary as the product of the
density and absolute temperature.
Ex. 1. The area of a cirde varies as the square of its radius,
and the area of a circle whose radius is 10 feet is 314*159 square
feet. What is the area of a circle whose radius is 7 f eet?
Ans, VHHii feet.
Ex. 2. The volume of a sphere varies as the cube of its radius,
and the volume of a sphere whose radius is 1 foot is 4*188 cubic feet.
What is the volume of a sphere of one yard radius? Am, 113*076 feet.
Ex. 3. The distance through which a heavy body falls from rest
varies as the square of the time it falls ; also a body falls 64 feet in
2 seconds. How far does a body fall in 6 seconds? Ana, 576 feet.
Ex. 4. The volume of a gas varies as the absolute temperature
and inversely as the pressure; also when the pressure is 15 and the
temperature 260 the volume is 200 cubic inches. What will the
volume be when the pressure becomes 18 and the temperature 390?
Arts, 250 inches.
250 INDETERMINATE FORMS.
Ex. 5. The distonce of the offing at sea Yftries as the sqaare root
of the height of the eye above the sea level, and the distance is
3 miles when the height is 6 feet: find the distance when the height
is 72 yards. Aru, 18 miles.
Indeterminate Forms.
217. A ratio or fraction sometimes assumes an in
determinate form for some value or values of a contained
letter.
Thus, when x=xO both the numerator and the denominator of
the fraction r vanish, and the fraction assumes for this value of
sr — X
X the indeterminate form tt ; and this is also the case when «= 1.
Again, when x=ao both the numerator and the denominator of
the above fraction become infinitely great, and the fraction assumes
the indeterminate form — •
00
We proceed to shew how to find the limiting values
of fractions which assume these indeterminate forms.
a? — 1
Consider, for example, the fraction 3 — ^, which as
sumes the form ^ when a?= 1.
^i_ (g^iK^+i) .
and, provided x — 1 is not really zero, we may divide the
numerator and denominator by a; — 1 without altering the
value of the fraction, and we can do this however small
X— 1 may be.
Hence, when a? — 1 is very smalL = — = = 5— —=• ,
and the limiting value of the latter fraction, as x ap
2
preaches indefinitely near to 1, is at once seen to be ^ .
INDETERMINATE FORMS. 251
Hence, as x approaches indefinitely near to 1, the
a^— 1 . . 2
fraction ,^ , approaches indefinitely near to the value ^.
a? — l 2
This is expressed by the notation X,„, Ji~Zr{ "^ ^ "^
x^ — 5x + 6
Ex. 1. Find the limiting value of « — Tjr :r^ when x=2,
X* — iMx + lo
It follows from Art. 88 that «2 ia a common factor of the
numerator and denominator.
a;'6a; + 6 _^ (a;~2)(g3) _ a;3_l
^«=2 «2_i0a.^.i6^«2 (a?2)(a;8)" *"• a;8~6'
a;*+2a;
Ex. 2. Find the limiting value of ^r^ — ^ when a;=0 and when
«2+2x_ a;(j? + 2) a; + 2 _2
*•"<> 2a?3 + 3x~ *"° a;(2a; + 3j'" *"* 2x + 3~3
(1) ,
2 3
since  and  are hoth zero when x is infinite.
x X
Ex. 3. Find the limiting value of the ratio l + 2:i; : 2 + 3a; when x
increases without limit.
^«— 2 + 3a?""*'*"* /„ 2\ *"• Q 2 3
3 + —
2a:2 I IQOx 4 500
Ex. 4. Find the limitmg value of Z T^ Z when a; becomes
oar 40
indefinitely great.
2a;g+100g + 500 r ^^^__^
"(^^^f)
^s)
_r ?f!_r 10
252 EXAMPLES.
EXAMPLES XX.
1. Shew that, if a + 6, 6 + c, c + a are in continued propor
tion, then b + c:c + a = c — a:a—b.
2. Shew that, ii xi a = y: b = z : c, then
3. Shew that, if(a + 6 + c + c?)(a6c + c?)=(a  6 + c  c?)
(a + 6 — c  rf), then a, 6, c, d are proportionals.
4. Shew that, if 6* + c* = a^ then
a + 6 + c : c + a — 6 = a + 6c: 6 + c— a.
6. What number must be subtracted from each of the
numbers 7, 10, 19, 31 in order that the remainders may be in
proportion?
6. Find a: b : c, having given
6 a + cb _ a + 6 + c ,
a + b b + c  a~^ 2a + b + 2c' \
7. K « _ y _ « I
b + ca c + a6 a\b — c^
shew that {a^b^ c) (yz + 2;aj + xy) = {x + y + z) (ax ^by •¥ cz),
8. If a (y + «) = 6 (« + a:) = c (a; + y), prove that
y—z z—x x— y
a{b — c) b {c a) c{ab)'
9. Shew that the ratio
la +1 a +1 a + : I b + lb +lb +
*'l*^l ' 8 8 ^^ 8 3 ••••••• "j^i 8 8 8 3
is intermediate to the greatest and least of the ratios a^: b^,
a^ib^, <fec., the quantities being all positive.
10. If a:b::c :df then
a^'^ + b^ + c'^ + d'^ _
11. Shewthat, if (a + 6)(6 + c) (c + c^ (ci + a)
= (a + 6 + c + CI?) (6cci? + cda + c£a6 + a5c),
then a : 6 : : d? : c.
EXAMPLES. 253
12. If (bed + cda ^ dab + abc)'  abed (a i b +e + dy=0,
then it will be possible to arrange Oj 6, c, c? so as to be propor
tionals.
13. Shew that, if — ^ — =i^ = ^— ,
a + 2o + c a^e a — 26 + c
then
x+2j/ + z x^z x2yiz'
14. Shew that, if ax^ + by' + e7^ + 2/yz + 2gzx + 2/wjy =
and x + y + z = are only satisfied by one set of ratios x \y\z,
then be /' + ca^ + a6A* + 2 (gh  a/)
+ 2{hf^bg)^2{fgch) = 0.
15. Shew that, if
a b e
p(pxqyrz) q(qy — rzpx) r{rz — px — qy)^
then y ^ g ^ ^
a (aa5  6y — cs;) 6 (Jyy — cz — ax) e{cz — ax^ by)
16. Shew that, if ah = cd, then either of them is equal to
(a + c) (a + cQ (6 + c) (6 + c?)/(a + 6 + c + dy.
Also, if a + 6 = c + c?, then either of them is equal to
ahed (~ + r + " *♦■ j) /C*^^ + c^) f A o tC^^^
17. Find the limiting values of the following fractions
when oj = 2, and when aj = oo .
aj'Oaj+U' ^^iB«5aj+6' ^ ^ ./r*12a;+ 16*
18. Find the limiting values of the following when 03 = e»,
^^ ija^x' W ^(a;«^a«) '
VnOt^ ^v
CHAPTER XVII.
r Arithmetical,K}eometrical, and Harmonical
Progression.
218. Series. A succession of quantities the members
of which are formed in order according to some definite
law is called a series.
Thus 1, 2, 3, 4, , in which each term exceeds the
preceding by unity, is a series.
So also 3, 6, 12, 24, , in which each term is double
the preceding, is a series. .
We shall in the present Chapter consider some very
simple cases of series, and shall return to the subject in a
subsequent Chapter.
Arithmetical Progression.
219. Definition. A series of quantities is said to be
in Arithmetical Progression when the difiference between
any term and the preceding one is the same throughout
the series.
Thus, a, b, c, d, &c. are in Arithmetical Progression
[a. P.] if 6 — a = c — 6 = d — c = &a
The difference between each tenn of an A. P. and the
preceding term is called the common difference.
ARITHMETICAL PBOGBESSION. 255
Hie foQowing are examples of Arithmetical {icpgreeBioiis: —
1, 3, 6, 7, &o.
8, 1, 6, 9, Ac.
a, a + 2b, a +46, &o.
In the first series the common difference is 2, in the second it is
" 4, and in the last it is 2b,
220. If the first term of an arithmetical progression
be a, and the common diflference d; then, by definition,
the 2nd term will he a + d,
„ 8rd „ „ af2d,
„ 4th „ „ af3d,
and so on, the coefficient of d being always less by unity
than the number giving the position of the term in the
series.
Hence the nth term will be a + (n — 1) d.
We can therefore write down any term of an A. P.
when the first term and the common diflference are given.
For example, in the a. p. whose first term is 5, and whose
common difference is 4, the 10th term is 5 + (10 1)4 =41, and the
30th term is 5+29x4=121.
221. An arithmetical progression is determined when
any two of its terms are given.
For, suppose we know that the mth term is or, and that
the nth term is )8.
Let a be the first term, and d the common diflference;
then the mth term will be a + (*/i  1) d, and the wth term
will be a f (ri — 1) d.
Hence a + (m — l)d = a,
and a + {n — l)d=)8.
Thus we have two equations of the first degree to
determine a and d in terms of the known quantities m, n,
a and fi.
256 ARITHMETICAL PROGRESSION.
Ex. Find the 10th term of the a. p. whose 7th tenn is 15 and whose
2l8t term is 22.
If a be the first term, and d be the common difference, we have
a+6t2=15, and a+20d=22.
1 9
Hence d=^ , a=12. The 10th term is therefore 12 + =16J.
222. When three quantities are in arithmetical pro
gression, the middle one is called the Arithmetic Mean of
the other two.
If a, 6, c are in A. P., we have, by definition,
6 — a = c — &; and therefore 6 = ^ (a + c).
Thus the arithmetic mean of two given quantities is half
their sv/m.
When any number of quantities are in arithmetical
progression all the intermediate terms may be called
aHthm^tio m£a/ns of the two extreme terms.
Between any two given quantities any number of arith
metic means may be inserted.
Let a and b be the two given quantities, and let n be
the number of terms to be inserted.
Then b will be the n + 2th term of the A. P. whose first
term is a.
Hence, if d be the common difference, 6 = a + (n + l)d;
and therefore d = — ^.
n + 1
Then the series is
6 — a ^b — a „
the required arithmetic means being
, b a . ckb — a b — a
a+ — — =, a + 2 — TT' » + ^ — TT*
n\\ nfl n + 1
na46 (ti  1) g + 26 (7i2)a + 36 a\nb
^^ n + 1' n + l ' n+l ' ' 1mT *
ARITHMETICAL PROGRESSION. 257
223. To find the sum of any nv/mber of terms of an
arithmetical progression.
Let a be the first term and d the common difiference.
Let n be the number of the terms whose sum is required,
and let I be the last of them.
Then, since I is the rith term, we have
? = a + (7il)d (i).
Hence, if ^ be the required sum,
fif=a + (a + d) + (a + 2d) + +(Z2d) + (?d) + l
Now write the series in the reverse order ; then
^ = Z + (?cZ) + (?2d) + ......+(a + 2d) + (a + d) + a.
Hence, by addition of corresponding terms, we have
2^ = (a + + (a + + (a + + ^0 n terms
.^.8=\{a+l) (ii),
or, from (i),
ig = {2ah(nl)d} (iii).
From the formulae (i), (ii), (iii) the value of all the
quantities a, d, n, I, 8 can be found when any three are
givea
Ex. 1. Find the snm of 20 terms of the arithmetical progression
3+6 + 9+&C.
Here a=3, d=3, w=20;
/. flf=^{6 + 19x3}=630.
4
Ex. 2. Shew that the sum of any number of consecutive odd numbers,
beginning with unity, is a square number.
The series of odd numbers is
1+3+6+
Here a=l, d=2; hence the sum of n terms is given by
8=^{2a + {nl)d}='^{2 + {nl)2}=n^
S. A. 17
258 ARITHMETICAL PROGRESSION.
Ex. 3. How many tenns of the series 1+5 + 9 + mnst be taken
in order that the sum may be 190?
We have /Sf= {2a + (n 1) d}, where 8=190^ a=l, c!=4.
Hence n is to be found from the quadratic equation
190={2+4(nl)},
or 2w»n190=0,
that is (n  10) (2n + 19) = 0.
19
Hence n=10. The value w= ^ is to be rejected for n must
necessarily be & positive integer*,
Ex. 4. How many terms of the series 5 + 7 + 9+ must be taken
in order that the sum may be 480?
Here we have
480={10 + (nl)2};
/. n2+4n 480=0,
or (n20)(n+24)=0.
Hence n must be 20, for the value n=  24 must be rejected as a
negative number of terms is altogether meaningless*.
Ex. 5. What is the 14th term of the a. p. whose 5th term is 11 and
whose 9th term is 7? Am. 2.
Ex. 6. What is the 2nd term of the a. p. whose 4th term is 6 and
whose 7th term is 3a + 46 ? Atu,  2a  6.
Ex. 7. Which term of the series 5, 8, 11, &q, is 320?
Ant. The 106th.
Ex. 8. Shew that, if the same quantity be added to every term of an
A.P., the sums wUl be in a.p.
Ex. 9. Shew that, if every term of an a. p. be multiplied by the same
quantity, the products will be in a. p.
* The inadmissible value is a root of the equation to which the
problem leads, but it is not a solution of the problem. [See Chapter xi.]
It should be remarked that a negative value of n cannot mean a number
of terms reckoned bachwards.
QEOMETRICAL PROGRESSION. 259
Ex. 10. Shew that, if between every two conseoutiye terms of an
A. p., a fixed number of arithmetic means be inserted, the whole will
form an arithmetical progression.
Ex. 11. Find the sum of the following series:
(i) 2J+4i+6i+ to 23 terms.
(ii) 2 "•"6 "6" to 12 terms.
(iii) (a+96) + (a + 76) + (a+66) + to 10 terms.
.. . n1 . n2 . n3 . . ,
(it) + + + ton terms.
^ ' n n n
Am. (i) 621, (u) 16, (iii) 10a, (iv) (nl).
Ex. 12. The 7th term of an a. p. is 15, and the 21st term is 8; find
tiie sum of the first 13 terms. Aiu. 195.
Ex. 13. Find the sum of 21 terms of an a. p. whose 11th term is 20.
Am. 420.
Ex. 14. Shew that, if any odd nmnber of quantities are in a. p., the
first, the middle and the last are in a. p.
Ex. 15. Shew that, if unity be added to the sum of any number of
terms of the series 8, 16, 24, <fec., the result will be the square of an
odd number.
Ex. 16. How many terms of the series 15 + 11+7 + must be
taken in order that the sum may be 35? Am. 5.
Ex. 17. The sum of 5 terms of an A. p. is  5, and the 6th term is
 13; what is the common difference? Am.  4.
Ex. 18. Find the sum of all the numbers between 200 and 400 which
are diyisible by 7. Am. 8729.
Ex. 19. If a series of terms in a. p. be collected into groups of n terms,
and the terms in each group be added together, the results form an
A. p. whose common difference is to the original common difference as
QEOMETRICAL PROGRESSION.
224. Definition. A series of quantities is said to be
in Oeometrical Progression when the ratio of any term
to the preceding one is the same throughout the series.
17—2
260 GEOMETBIOAL PBOGBESSION.
Thus a, 6, c, d, &c. are in Geometrical Progression
(g.p.) if = = =  = &c.
Qf '
The ratio of each term of a geometrical progression
to the preceding term is called the common ratio.
The following are examples of geometrical progressions:
1, 3, 9, 27, &o.
4, 2, 1, J, &c.
a, a', aJ^, o^, <fec.
In the first series the common ratio is 3, in the second series it is
 \t and in the third series it is a\
225. If the first term of a G.P. be a, and the common
ratio r; then, by definition,
the 2nd term will be ar,
„ 3rd „ „ ar^,
„ 4th „ „ ar^,
and so on, the index of r being always less by unity than
the number giving the position of the term in the series.
Hence the nth term will be ar*"\
We can therefore write down any term of a G.P. when
the first term and the common ratio are given.
For example, in the o.p. whose first term is 2, and whose common
ratio is 3, the 6th term is 2 x 3», and the 20th term is 2 x 3^*.
226. A Geometrical Progression is determined when
any two of its terms are given.
For, suppose we know that the mth term is a, and
that the nth term is fi.
Let a be the first term, and r the common ratio;
then the mth term will be ar'^'\ and the nth term will
be ar^'K
GEOMETBICAL PKOGRESSION. 261
Hence ar"^'"^ = a, ar''''^ = ^8 ; and /. r*""* = ~ .
P
1 1 ln 1TO
Hence r = a"»"»y8**"*», and therefore a= a"»"»^~~'*.
Ex. Find the first term of the a. p. whose 3rd term is 18 and whose
5th term is 40^.
If a be the first term, and r the common ratio, we have
ar»=18, ar^=^%; 'r'^X*
2 4
Hence 0=18x5=8.
Thus the series is 8, 12, 18, &o,
227. When three quantities are in G.P., the middle
one is called the Geometric Mean of the other two.
If a, b, c are in G.P., we have by definition
be L . /
a ^
Thus the geometric mean of two given quantities is
a square root of their product.
When any number of quantities are in geometrical
progression all the intermediate terms may be called
geometric m£ans of the two extreme terms.
Between two given quamtities any number of geometric
means may be inserted.
For let a and b be the two given quantities, and let n
be the number of means to be inserted.
Then b will be the (w + 2)th term of a G.P. of which a
is the first term. Hence, if r be the common ratio, we
have
n+l/h
V a
Hence the required means are ar, ar^, .ar"^,
n 1 n1 2 1 n
that is, a^^+ift'^+i, a'*+i6'*+S a'*+i6^+i.
262 GEOMETRICAL PBOGBESSION.
228. To find the sum of any nv/mber of terms in
geometrical progression.
Let a be the first term, and r the common ratio. Let
n be the number of the terms whose simi is required, and
let I be the last of them.
Then, since I is the nth term, we have I = ar"*^.
Hence, if fif be the required sum,
fif=a + ar + a7^+ +ar'*"\
Multiply by r ; then
Sr= ar + ar^^ ar*\ + ar""'^ + ar^»
Hence, by subtraction,
8^ Sr=sa — ar* ;
1 — r*
.". 8=az: .
1— r
Ex. 1. Find the sum of 10 terms of the series 3, 6, 12, &o.
Here a=8, r=2, n=10.
Hence fif=3if^*=3 (2^ 1)=3069.
229. From the preceding article we have
1 ~ r'* a ar"^
S = a
1— r 1— r 1
Now when r is a proper fraction, whether positive or nega
tive, the absolute value of 7^ will decrease as n increases ;
moreover the value of r* can be made a^ small as we
please by sufficiently increasing the value of n.
Hence, when r is nwmerically less than unity, the sum
of the series can be made to differ from :; — by as small
a quantity as we please by taking a sufficient number of
terms.
QBOMETRIOAL PBOQBESSION. 263
Thus the sum of an infinite number of terms of the
geometrical progression a + ar + ar^ + , in which r is
CL
nu/merically less than tmity, is = .
Ex. 1. Find the sum of an infinite number of tenns of the series
96+4
Here a=9, r=5.
o
TT « a 9 27
Henoe S=^ — = ' / ov =e'
1r  / 2\ 6
(i)
Ex. 2. Find the geometrical progression whose snm to infinity is H,
and whose second term is  2.
Let a be the first term, and r be the common ratio. ^
Then we have arss  2, and = — = ^ •
1r 2
Whence 9r»9r4=0.
Henoe — g. <=.
If r=i,a=:i?=65
2
and the series is 6,  2, ^ , &o.
8
4. .
The valne r =^ is inadmissible, for r must be nomerioally less than
o
unity.
Ex. 8. The 3rd term of a o.p. is 2, and the 6th term is  J ; what is
the 10th term? Am, if
Ex. 4. Insert two geometric means between 8 and 1, and three
means between 2 and 18. Ans, 4, 2; =t 2/^3, 6, st 6/^3.
Ex. 5. Shew that if all the terms of a o.p. be multiplied by the same
quantity, the products will be in a. p.
Ex. 6. Shew that the reciprocals of the terms of a o.p. are also in a.p.
Ex. 7. Shew that, if between every two consecutive terms of a o.p., a
fixed number of geometric means be inserted, the whole will form a
geometrical progression.
264 HARMONICAL PBOOBESSION.
Ex. 8. Find the sum of the following series :
(i) 12 + 9 + 6i+... to20tennB.
2 4
(ii) lg + Q+»»« to6tenns.
(iii) 4+8 + 16+... to infinity.
^n..(i)48{lg)'"j., (ii)g, (iii) 5.
Ex. 9. Shew that the continued product of any number of quantities
n
in geometrical progression is equal to (^Z)^, where n is the number of
the quantities and Qj I are the greatest and least of them.
Ex. 10. Shew that the product of any odd number of terms of a a.p.
will be equal to the nth power of the middle term, n being the number
of the terms.
Ex* JLl. The sum of the first 10 terms of a certain o.p. is equal to 244
times the sum of the first 5 terms. What is the common ratio ?
Ans, 3.
Ex. 12. If the common ratio St a g.p. be less than }, shew that each
term will be greater than the sum of all that follow it*
HARMONICAL PROGRESSION.
230. Definition. A series of quantities is said to be
in Harmonical Progression when the difference between
the first and the second of any three consecutive terms is
to the difference between the second and the third as the
first is to the third.
Thus a, b, c, d &c., are in Harmonical Progression
[h. p.], if
a — b : b — o :: a : c,
b — c:c — d::b:d,
and so on.
If a, b, c be in harmonical progression, we have by
definition
a — 6:6 — c::a:c;
,% c (a — 6) = a (6  c).
HARMONICAL PROGRESSION. 265
Hence, dividing by ahc, we have
Thus, if quantities are in harmonical progression, their
reciprocals are in arithmetical progression.
231. Harmonic Mean. If a, b, c be in harmonical
111...
progression, , r,  will be in arithmetical progression.
Hence r "= ~ +  >
a c
J 2ac
a^c
Thus the harmonic mean of two quantities is twice
their product divided by their sum.
If we put A, 0, H for the arithmetic, the geometric,
and the harmonic means respectively of any two quantities
a and 6, we have
.'.A.H^CP.
Thus the geometric mean of any two quantities is also
the geometric mean of their arithmetic and harmonic
means,
232. Theorem. ITie arithmetic mean of two uneqvxil
positive quantities is greater than their geometric mean.
If a, b be the two positive quantities we have to shew
that
i (a + 6) > Job,
or i (Va  ^/by > 0.
266 EXAMPLES.
Now {hja — isjhy is always positive, and therefore greater
than zero, unless a = 6.
Since the arithmetic mean of two positive quantities is
greater than their geometric mean, it follows from Art. 231
that the geometric mean is greater than the harmonic.
233. To insert n harmonic means between any two
quantities a and 6.
Insert n arithmetic means between i and J, and the
a
reciprocals of these will be the required harmonic means.
The arithmetic means are
1 J_/l_l\ 1 2 /I 1\ o
Hence, by simplifying these terms and inverting them,
the required harmonic means will be found to be
(n + 1) ah (n + 1) ab (n + 1) ab
nb + a '(nl)6 + 2a' ' b + na *
234. It is of importance to notice that no formula can
be found which will give the sum of any number of terms
in harmonical progression.
EXAMPLES XXI.
1. Shew that, if a, 6, c be in a. p., then will a^(h + e)y
b' (c + a), c* (a + 6) be in A. P.
2. Find four numbers in A. P. such that the sum of their
squares shall be 120, and that the product of the first and last
shall be less than the product of the other two by 8.
3. If a, 5^ c be in a. p., and 6, c, c? be in h.p., then will
a:b = c:cL
4. Find three numbers in g.p. such that their sum is 14,
and the sum of their squares 84.
EXAMPLES. 267
5. If a, 6, c be in arithmetical progression, and x be the
geometric mean of a and b, and y be the geometric mean of b
and c ; then will a^, b\ y" be in arithmetical progression.
6. Shew that, if a, 5, c be in harmonical progression, then
will ^ , y and — , be also in harmonical
b+ca c+ab a+b—c
progression.
7. Shew that, if a, b, c, d he in harmonical progression,
then will
3(ba){d^c)=:{c^b){da).
8. Shew that, if a, &, c be in harmonical progression^
b b — a b^c'
9. Shew that, if a, 6, c be in h.p., then will
b+a b+c
b—a 6— c
= 2.
10. If a, b, che in a. p., b, Cy din. o. p., and c, d, e in h. p. ;
then will a, c, ehe in g. p.
11. If a, 6, c be in H. P., then will a  h > h > <^ "" h ^® "^ ®p*
12. If a, 6, c are in h.p., then a^ a — c, a — b are in H.P.,
and also c, c — ay c — b are in h. p.
13. If fic, ttj, a^y y be in a. p., a;, g^y g^y y in g.p., and
Xyh^yh^y yia. H.P., then
AjAj ^1 + A, *
14. The sum of the first, second, and third terms of a G. P.
is to the sum of the third, fourth and fifth terms as 1 : 4, and
the seventh term is 384. Find the series.
268 EXAMPLES.
15. If a,, a„ a., ,a be in harmonical progression,
prove that a^a^ + ajx^ + a^a^ + ..".... + »._,«, = (w  1) a^a^.
16. If a, 05, y, 6 be in arithmetical progression, and
a, Uy V, & be in harmonical progression, then qcv= yu = ah,
17. Three numbers are in arithmetic progression, and the
product of the extremes is 5 times the mean ; also the sum of
the two largest is 8 times the least. Find the numbers.
18. If = = , 6, z — = be in a. p. : then a, 7 , c will be
1 — a616c
in H.P.
19. If a, 6, c be in A, p., and a', 5", c* be in H.P., prove
that — ^ , 6, c are in G. p., or else a = 6 = c.
20. If X be any term of the arithmetical progression and y
be the corresponding term of the harmonical progression whose
first two terms are a, 6, then will x—a : y — a :: b : y.
21. Shew that, if a be the arithmetic mean between h and
c, and h be the geometric mean between a and c, then will c be
the harmonic mean between a and h.
22. The series of natural numbers is divided into groups as
follows: 1; 2, 3; 4, 5, 6; 7, 8, 9, 10; and so on. Prove that
the sum of the numbers in the k^^ group is ^k (^ + 1).
23. An A. p. and an h. p. have each the fii*st term a, the
same last term I, and the same number of terms n ; prove that
the product of the (r + 1)*** term of the one series and the
(n — r)*^ term of the other is independent of r,
24. Terms equidistant from a given term of an A. p. are
multiplied together ; shew that the differences of the successive
terms of the series so formed are in a.p.
25. Shew that, if S^, S^^y S^^ be the sum of n terms, of
2/1 terms, and of 3n terms respectively of any q.p., then will
26. If a, b, c be all positive and either in a.p., in g.p.,
or in H. p., and n be any positive integer, then a" + c" > 25".
EXAMPLES. 269
27. If P, Q, R be respectively the p^^ q^, and r^ terms
(i) of an A. p., (ii) of a G. p., and (iii) of an h. p., then will
(i) F(qr) + Q(rp)^B{pq) = 0,
(ii) P»'.e''"'.^~' = l,
(iii) QB(qr) + RP{rp) + PQ{p.q)^0.
28. Shew that, if CTj, Ojj %> > <*„ ^^ i^ h.p., then
^i ^« «,
will be in H. p.
29. Shew that, if a^, a^, ©g , a^ be all real, and if
« + <+ + <i)K' + <+ +0
then will a^, a,, ag, be in G. P.
30. Shew that any even square, {^rCf, is equal to the sum
of n terms of one series of integers in a. p., and that any odd
square, {2n + 1)', is equal to the sum of n terms of another a. p.
increased by unity.
31. Prove that any positive integral power (except the
first) of any positive integer, p, is the sum of p consecutive
terms of the series 1, 3, 5, 7, &c. ; and find the first of the p
terms when the sum is p",
32. If an A. p. and a a p. have the same first term and the
same second term, every other term of the A. p. will be less
than the corresponding term of the g.p., the terms being all
positive.
CHAPTER XVIII.
Systems op Numeration.
235. In arithmetic any number whatever is repre
sented by one or more of the ten symbols 0, 1, 2, 8, 4, 6, 6,
7, 8, 9, called figures or digits, by means of the convention
that every figure placed to the left of another represents
ten times as much as if it were in the place of that other.
The cipher, 0, which stands for nothing, is necessary
because one or more of the denominations, units, tens,
hundreds, &c., may be wanting.
The above mode of representing numbers is called the
common scale of notation, and 10 is said to be the radix or
base.
236. Instead of ten any other number might be used
as the base of a System of NvmeraMon, that is of a system
by which numbers are named according to some definite
plan, and of the corresponding Scale cf Notation, that is
of a system by which numbers are represented by a few
signs according to some definite plan ; and to express a
number, N, in the scale whose radix is r, is to write the
number in the form ...... djclji^d^, where each of the digits
^o» ^i> ^8> ^8 ^ l^ss than r, and where d^^ stands for d^
units, dj stands for d^ x r, d, for d, x r^, and so on.
Thus N=^d^ + d^r+dy+
Note. Throughout this chapter each letter stands for a
positive integer, unless the contrary is stated.
SYSTEMS OF NUMERATION. 271
237. Theorem. Any positive integer can he expressed
in any scale of notation, and this can be done in only one
way.
For divide N by r, and let Q^ be the quotient and d<,
the remainder.
Then N = d^ + rxQ^.
Now divide Q. by r, and let Q be the quotient and d.
the remainder.
Then Q^ = d^ + rQ, ; therefore N = d^ + rd^\ r*Q,.
By proceeding in this way we must sooner or later
come to a quotient, Qn = d^, which is less than r, when the
process is completed, and we have
JV=do + rdj + r'd, + r'd,4 r^d^,
so that the number would in the scale of r be written
dn djdjd^df^.
Each of the digits d^, d^, d^, is less than r, and any
one or more of them, except the last, d^, may be zero.
Since at every stage of the above process there is only
on^ quotient and one remainder the transformation is
unique.
The given number N may itself be expressed either in
the common or in any other scale of notation.
Ex. 1. Express 2157 in the scale of 6.
The quotients and remainders of the sncoessiye divisions by 6
are as under:
6 1 2157
6 1 359 remainder S=dQ
6I69 6=di
6[9 5=dj
1 3=(f3
Thus 2157 when expressed in the scale of 6 is 18553.
Ex. 2. Change 13553 from the scale of 6 to the scale of S.
We have the following successive divisions by 8, remembering
that since 13553 is in the scale of 6 each figure is six times what
it would be if it were moved one place to the left, so that to begin
with we have to divide 1x6 + 3, and so on.
272 SYSTEMS OF NUMERATION.
8 1 13553
8 [1125 remainder 5
863 5
4 1
Hence the number required is 4155.
Ex. 3. Change 4155 from the scale of 8 to the scale of 10.
Proceeding as before, we have
10 1 4155
10 1 327 remainder?
1025 5
2 1
Thus 2157 is the number required.
Or thus:
Since 4155=4 x8»+lx 82 + 5 x8 + 5={(4x8 + l)8 + 5}8+5,
the required result may be obtained as follows : —
Multiply 4 by 8 and add 1; multiply this result by 8 and add 5 ;
then multiply again by 8 and add 5.
Ex. 4. Express 3166 in the scale of 12. [Represent ten by t, and
eleven by e,] Ant. l^eU
Ex. 5. Express ^ in the scale of 4. Am. ^t? •
Ex.6. In what scale is 4950 written 20301? Ans. 7.
238. Radix Fractions. Radix fractions in any scale
correspond to decimal fractions in the ordinary scale, so that
'ahc... stands for  + ^ + ^+
r IT 'T
To shew that any given fraction may he expressed by a
series of radiw fractions in any proposed scale.
Let F be the given fraction; and suppose that, when
expressed by radix fractions in the scale of r, we have
r IT r
F=abc = 'f: + ^ + ^ +
where each of a, 6, c is a positive integer (including
zero) less than r.
SYSTEMS OF NUMERATION. 273
Multiply by r; then
Tj be
r IT
Hence a must be equal to the integral part, and
h c
 + 3 + must be equal to the fractional part of Fr.
r IT
(If Fr be less than 1, a is zero.)
Let F^ be the fractional part of Fr\ then
^.=^^+
r r
Multiply by r\ then
F xr=6++
1 ^
Hence h must be equal to the integral part of F^.
Thus a,b,Cy can be found in succession.
Ex. 1. Express ^hj a series of radix fractions in the scale of 6.
Hence '012 is the required result.
Ex. 2. Express = by a series of radix fractions in the scale of 3.
ix3 = 0+2; ?x3 = l+?; fx3 = 0+;
?x3=2+^; x3=l+; x3=2+i.
Hence >dl02lS is the required result.
Ex. 3. Change 324*26 from the scale 8 to the scale 6.
The integral and fractional parts must be considered separately.
S. A. 18
274 SYSTEMS OF NUMERATION.
61324
•26
6 1 43 remainder 2
5 5
6
204
6
030
6
220
6
140
6
300
Thus the required result is 55220213.
Ex. 4. In the scale of 8 express '16^15 as a vulgar fraction.
2^= 16315;
.. 8«2f= 16311;;
/. 8W=16316Sia;
16315  16 16316  16 16277
N.
8»8» 77700 77700
Ex. 5. In the scale of 7 express *23l as a vulgar fraction.
'*'"• 330'
Ex. 6. Change 31423 from the scale of 5 to the scale of 7.
Ana, 150S56I.
239. Theorem. The difference between any nwmber
and the sum of its digits is divisible 6y r — 1, where r is the
radix of the scale in which the number is caressed.
Let N be the number, S the sum of the digits, and let
dQ, d^, dj be the digits.
Then N=^d^ + rd, + r^d^ + + r^.
and S^^d^ + d^ + d^h. + d^,
... JV'fif = (rl)d^ + (r*l)d,+ + (r^^l)d^.
Now each of the terms on the right is divisible by
r:l [Art. 86].
Hence JV" — fi^ is divisible by r — 1.
Since N—8 is divisible by r — 1, JV and 8 must leave
the same remainder when divided by r — 1.
SYSTEMS O^ NUMfiRATIOK. 275
Ex. 1. The difference of any two numbers expressed by the same
digits is divisible by r 1.
For the sum of the digits is the same for both; and since N^~8
and Nm8 are both divisible by r1, it follows that NiN^ is
diyisibfe by r  L
Ex. 2. Shew that in the ordinary scale a number is divisible by 9 if
the sum of its digits be divisible by 9, and by 3 if the sum of its digits
is divisible by 3.
^ £f is a multiple of 9 ; hence, if £» be a multiple of 9, so also is
N; and, if iS be a multiple of 3, so also is i^.
Ex. 3. Shew that any number is divisible by r+ 1 if the difference
between the sum of the odd and the sum of the even digits is
divisible by r+1.
Let N=d0+djrhd2r^+d^i^^ ,
and D=d^di+d^d^+
Then ^'D=di(r+l)+d,(r3l)+d,(r»+l) +
Each of the terms on the right is divisible by r+1 [Art. 87];
/. NDiB divisible by r + 1. Hence if D is divisible by r+ 1 so also
is 2^.
Ex. 4. If N^ and N^ be any two whole numbers, and if the remainders
left after dividing the sum of the digits in N^, N^ and in NixN^ by
9 be 94, n, and p respectively; then will 9491^ be equal to p, or diiffer
from > by a multiple of 9.
For 2^i=ni+a multiple of 9, and N=n^ + ik multiple of 9;
therefore N^x N^=niXru+ & multiple of 9. Hence njU^ + a multiple
of 9 is equal to jp + a multiple of 9.
If the above is applied in any case of multiplication, and it is
found that n^n, does not equal p, or differ from it by a multiple of 9,
there must be some error in the process of multiplication.
This gives a method of testing the accuracy of multiplication;
the test is not however a complete one, for although it is certain that
there must be an error if n^ x n^ does not equal p, or differ from it by
a multiple of 9, there fna^ oe errors when the condition is satisfied,
provided that the errors neutralize one another so far as the sum of
the digits in the product is concerned.
This is called the "Rule for casting out the nines."
Ex. 5. A number of three digits in the scale of 7 has the same digits
in reversed order when it is expressed in the scale of 9 : find the
number.
Let a, 6, c be the digits ; then we have
49a+76+c=81c+96+a,
where a, 5, c are positive integers less than 7>
Hence 40c + 6 = 24a.
18—2
276 EXAMPLES.
Now 40c and 24a are both divisible by 8; therefore b mnst be
divisible by 8. But b is less than 7 : it most therefore be zero. And
since b is zero, we have 5c=3a, which can only be satisfied when
c=3anda=5.
Thus the number required is 503.
Ex. 6. A nomber consisting of three digits is doubled by reversing the
digits; prove that the same will hold for the number formed by the
first and last digits, and also that such a number can be found in
only one scale of notation out of every three.
Let the number be abc in the scale of r.
Then we have {abc) x2=cba.
Since cba is greater than o&c, c must be greater than a.
Hence we must have the following equations :
2c=a+r (i),
26+l=6+r (ii),
2a + l=c (iii).
From (i) and (iii) we see that the number represented by ca is
double that represented by ac.
Also 4a+2=2c=a+r;
r2=3a.
• .
Hence, as a is an integer, r2 must be a multiple of 3, so that
the number must be in one of the scales 2, 5, 8, 11, <&o., the numbers
corresponding to these scales being Oil, 143, 275, 3t7, &o.
EXAMPLES XXn.
1. Find the number which has the same two digits when
expressed in the scales of 7 and 9.
2. In any given scale write down the greatest and the
least number which has a given number of digits.
3. A number of six digits is formed by writing down any
three digits and then repeating them in the same order ; shew
that the number is divisible by 1001.
4. Of the weights 1, 2, 4, 8, <&c. lbs., which must be taken
to weigh 1027 lbs. 1
EXAMPLES. 277
5. Shew that the number represented in any scale by 144
is a square number.
6. Shew that the numbers represented in any scale by
121, 12321, and 1234321 are perfect squares.
7. Find a number of two diirits, which are transposed by
the addition of 18 to the numb^ ^ by converting it into the
septenary scale.
8. A number is denoted by 4*440 in the quinary scale,
and by 4*54 in a certain other scale. What is the radix of
that other scale)
9. If aS^ be the sum of the digits of a number N^ and 2Q
be the sum of the digits of 2i\r, the number being expressed in
the ordinary scale, shew that ^^ ^ is a multiple of 9.
10. If a whole nimiber be expressed in a scale whose
radix is odd, the sum of the digits will be even if the number
be even, and odd if the number be odd.
11. Prove that, in any scale of notation, the difference of
the square of any number of three digits and the square of the
number formed by reversing the digits is divisible by r*— 1,
12. Prove that, in any scale of notation, the difference of
the square of any number and ilie square of the number formed
by reversing the digits is divisible by r* — 1.
13. A number of three digits in the scale of 7, when
expressed in the scale of 11 has the same digits in reversed
order : find the number.
14. Prove that all the numbers which are expressed in
the scales of 5 and 9 by using the same digits, whether in the
same order or in a different order, will leave the same remainder
when divided by 4.
15. There is a certain number which is expressed by 6
digits in the scale of 3, and by the last three of those digits in
the scale of 12. Find the number.
278 EXAMPLES.
16. Find a number of four digits in the scale of 8 which
when doubled will have the same digits in reverse order.
17. The digits of a number of three digits are in a. p.
The number when divided by the sum of its digits gives a
quotient 15 ; and when 396 is added to the number, the sum
has the same digits in inverted order. Find the number.
18. Find the digits a, 5, c in order that the number
13a&45c may be divisible by 792.
19. Prove that there is only one scale of notation in
which the number represented by 1155 is divisible by that
represented by 12, and find that scale.
20. Find a number of four digits in the ordinary scale
which will have its digits reversed in order by multiplying
by 9.
21. In the scale of notation whose radix is r, shew that
the number (r* — l)(r" — 1) when divided by r — 1 will give a
quotient with the same digits in the reverse order.
22. Shew that^ in any scale of notation,
the circulating period consisting of all the figures in order
except r — 2 which is passed over. For example, in the
ordinary scale, ^ = 01234567&.
23. There is a number of six digits such that when the
extreme lefthand digit is transposed to the extreme righthand,
the rest being unaltered, the number is increased threefold.
Prove that the lefthand digit must be either 1 or 2, and find
the number in either case.
24. Find a number of three digits, the last two of which
are alike, such that when multiplied by a certain number it
etill consists of three digits, the first two of which are alike
and the same as the former repeated ones, and the third is the
same as the multiplier.
CHAPTER XIX.
Permutations and Combinations.
240. Definition. The diflferent ways in which r
things can be taken from n things, regard being had to
the order of selection or arrangement, are called the per
mutations of the n things r at a time.
Thus two permutations will be diflferent unless they
contain the same objects arranged in the same order.
For example, suppose we have four objects, represented
by the letters a, 6, c, d; the permutations two at a time
are a6, ha, ac, ca, ad, da, be, cb, bd, db, cd, and dc.
The number of permutations of n diflferent things
taken r at a time is denoted by the symbol ^P^.
241. To find the number of permutations ofn different
things taken r at a time.
Let the diflferent things be represented by the letters
a, b, c,
It is obvious that there are n permutations of the n
things when taken one at a time, so that ^P^ = n.
Now in the permutations of the n letters r together,
the number of permutations in which a particular letter
occurs first in order is equal to the number of permuta
tions of the remaining n — 1 letters r — 1 at a time. This
is true for each one of the n letters, and therefore
280 PERMUTATIONS.
Since the above relation is true for all values of n
and r, we have in succession
But n^^,Px = (^^+l).
Multiply all the corresponding members of the above
equalities, and cancel all the common factors; we then
have
^P^ = n(nl)(n~2) (nr+l).
If all the n things are to be taken, r is equal to w, and
we have
^P^ = n(nl)(n2) 3.2.1.
Definitions. The product w (n  1) (n  2) . . . 2 . 1
is denoted by the symbol \n^ or hy n\ The symbols [n
and n\ are read ' factorial n.'
The continued product of the r quantities w, w — 1,
n — 2y (n — r+1), n not being necessarily an integer
in this case, is denoted by n^. Thus 7I3 = w (n — 1) (w — 2).
Hence we have P. = I w , and P^ = n.
242. To find the number of permutations of n things
taken all together, when the things are not all different
Let the n things be represented by letters; and sup
pose p of them to be a*s, q of them to be 6's, r of them to
be c*s, and so on. Let P be the required number of per
piutations,
PERMUTATIONS. 281
If in any one of the actual permutations we suppose
that the a's are all changed into p letters diflferent from
each other and from all the rest; then, by changing only
the arrangement of these p new letters, we should, instead
of a single permutation, have \p different permutations.
Hence, if the a's were all changed into p letters
different from each other and from all the rest, the Vs, c's,
&c. being unaltered, there would he P x\p permutations.
Similarly, if in any one of these new permutations we
suppose that the 6's are all changed into q letters different
from each other and from all the rest, we should obtain
q permutations by changing the order of thase q new
otters. Hence the whole number of permutations would
now be P X Ip X ly.
By proceeding in this way we see that if all the letters
were changed so that no two were alike, the total number
of permutations would he Px\p x \qx \r...
But the number of permutations all together of n
different things is \n. Hence P x ^ x l£X r...= \n;
"" In ~  ~
• p= !=
. \P\^\Z'"'
Ex. 1. Find gPs, gP^ and 7P7. Ans, 120, 120, 5040.
Ex. 2. Shew that ioi'4=7P7.
Ex. 3. If ^Pa = 12 X ^jPg, find n. Ans. 7.
Ex. 4. If j,jPg= 100 X ^Pa, find n. Ans, 13.
Ex. 6. If 2»P8= 2 X J?^, find w. Am. 8.
Ex. 6. Find the number of permutations of all the letters of each of
the words acacia, Tiannahf success and mtssissippu
Ans. 60, 90, 420, 34650.
Ex. 7. In how many ways may a party of 8 take their places at a
round table; and in how many ways can 8 different beads be strung
on a necklace ? Ans. 7, i [7.
Ex. 8. In how many ways may a party of 4 ladies and 4 gentlemen be
arranged at a round table, the ladies and gentlemen being placed
alternately? ^ Ans. 144,
282 COMBINATIONS.
Ex. 9. The number of permutations of n things all together in which
r specified things are to be in an assigned order though not necessarily
consecutive is w/jr.
Ex. 10. The number of ways in which n books can be arranged on a
shelf so that two particular books shall not be together is (n ~ 2)  nl.
Ex. 11. Find the number of permutations of n thingsr together, when
each thing can be repeated any number of times.
Here any one of the n things can be put in the first place; and,
however the first place is filled, any one of the n things can be put in
the second place; and so on. Hence the number required
=nxnxnx ... = /t^
Combinations.
243. Definition. The diflferent ways in which a
selection of r things can be made from n things, without
regard to the order of selection or arrangement, are called
the cornhinations of the n things r at a time.
Thus the diflFerent combinations of the letters a, b, c, d
three at a time are abc, abd, acd and bed.
The number of combinations of n different things r at
a time is denoted by the symbol JJ^.
244. To find the number of combinations ofn different
things taken r at a time.
Let the different things be represented by the letters
a^ Of c, • • •
Now in the combinations of the n letters r together
the number in which a particular letter occurs is equal to
the number of combinations of the remaining n — 1 letters
r — 1 at a time. Hence in the whole number of combina
tions r together every letter occurs ^.^G^^ times, and
therefore the total number of letters is n x ^.^0^^ ; but,
since there are r letters in each combination, the total
number of the letters must be r x O.
COMBINATIONS. 283
Hence rxJJ^^nx ^fir^t*
Since the above relation is true for all values of n and
of r, we have in succession
(r  1) X ^_fi^^ = (ri  1) X ^JJ,^,
(r2)x,_,(7^,==(n2)x„.30,.3,
2 X „>^A = (^^ + 2) X ^^fi^.
Also nr+A = ^^+l
Hence, by multiplying corresponding members of the
above equations and cancelling the common factors, we
have
r x^(7^=n(wl)(n2) (nr + 1),
thatis,a=^(^^>(^"y(^^ + ^> = ^ (i).
By multiplying the numerator and denominator of the
fraction on the right by In — r, we have
n (w  1) (n — 2). . .(n — r + 1) X \n — r
'^ Ir 71 — r
\n
(")•
[r \n — r
By comparing the above result with that obtained in
Art. 242, it will be seen that ^P^ = jO, x [r. The relation
JP^ = Jj^ x [r can however be at once obtained from the
consideration that every combination of r different things
would give rise to r permutations, if the order of the
letters were altered in every possible way.
Note. In order that the formula (ii) may be true
when r = n, we must suppose that 0 = 1, since ^(7^ = 1.
We should also obtain the result 0 = 1 by putting » = 1 in
the relation \n = n^ [ti— 1.
284 COMBINATIONS.
245. Theorem. The number of combinations of n
different things r together is equal to the number of the
combinations n — r together.
The proposition follows at once from the fact that
whenever r things are taken out of n things, n — r must
be left, and if every set of r things differs in some par
ticular from every other, the corresponding set of n — r
things will also differ in some particular from every other ;
and therefore the number of different ways of taking r
things must be just the same as the number of different
ways of leaving or taking n — r things.
The result can also be obtained from the formula
(ii) of the last Article.
For (7, = — ;^ — , and «CL^= ■ . ,
n — r \r
It should be remarked that the first method of proof
holds good when the n things are not all different, to which
case the formulae of Art. 244 are not applicable.
Ex. 1. Find ^qG^, ^C^ and 30^17. Ans. 210, 220, 1140.
Ex. 2. If n<76=n^ia» fi^d n<^16« ^^' l^^.
Ex. 3. Find n, having given that n^s—tfie ^^' !!•
Ex. 4. Find n, having given that 3 x ^O^ = 6 x njCj . Ans, 10.
Ex. 6. Find w, having given that „C4=210. Ans, 10.
Ex. 6. Find n and r having given that „P^= 272 and »Cy = 136.
Ans, n=17, rss2,
Ex. 7. Find n and r having given n^ri : n^r • n^r+i "2:3:4.
Ans, »=34, r=14.
Ex. 8. How many words each containing 3 consonants and 2 vowels
can be formed from 6 consonants and 4 vowels ?
The consonants can be chosen in ^C^=20 wa3r8; the vowels can be
chosen in 4(73=: 6 ways ; hence 20 x 6 different sets of letters can be
chosen, and each of these sets can be arranged in qPq=120 ways.
Hence the required number is 20 x 6 x 120,
COMBINATIONS. 285
Ex. 9. How many different stuns can be formed with a sovereign, a
halfsovereign, a crown, a halfcrown, a shilling and a sixpence ?
Nmnber required ^eCi+^C^+^C^+^C^+^C^+^Cg^QS.
Ex. 10. Shew that, in the combinations of 2n different things n
together, the nmnber of combinations in which a particular thing
occurs is equal to the number in which it does not occur.
Ex. 11. Shew that, in the combinations of 4n different things n
together, the number of combinations in which a particular thing
occurs is equal to onethird of the number in which it does not
occur.
Ex. 12. Out of a party of 4 ladies and 8 gentlemen one game at lawn
tennis is to be arranged, each side consisting of one lady and one
gentleman. In how many ways can this be done ? Ans. 36.
Ex. 13. The figures 1, 2, 8, 4, 6 are written down in every possible
order : how many of the numbers so formed will be greater than 23000 ?
Ans. 90.
Ex. 14. At an election there are four candidates and three members
to be elected, and an elector may vote for any number of candidates
not greater than the number to be elected. Li how many ways may
an Sector vote? Ans, 14.
246. Greatest value of jO^ To find the greatest
value of ^G^ for a given value ofn.
From the formulae of Art. 244 we have
Hence JJ^ = tiPr^v according as n — r + 1 r\ that is,
according as r = (n + 1).
Thus the number of combination of n things r together
increases with r so long as r is less than ^ (n + 1).
n
If then n be even, JJ^ is greatest when ^ = s •
If « be odd, nOr^nOr^i as r>^(n + l), and ,(7^ =
JO^^^ when r = J (n + 1). Thus, when n is odd,
^(7j^^jj = ^(7j(^^^, and these are the greatest values of „(7^.
286 COMBINATIONS.
For example, if n=10, n^r ^^ greatest when r=5. Also if nslly
JCy, is the same for the values 5 and 6 of r, and ffi^\A greater for these
yalues than for any other value.
247. To p'ove that JJ^ + JJ^^^ = ^fi^.
If the total number of combinations of (n + 1) things r
together be divided into two groups according as they do or
do not contain a certain particular letter, it is clear that the
number of the combinations which do not contain the
letter is the number of combinations r together of the
remaining n things, and the number of the combinations
which do contain the letter is the number of ways in which
r — 1 of the remaining n things can be taken. Thus
The above result can also be proved as follows :
From Art. 244 we have
^ _n(n — l)...(n — r + 1) n(n— l)...(n — r + 2)
" •• "*" "^•' " 1.2 r ^ 1.2 (r1) ~
n(n — 1)... (n — r + 2) f ^ >
""^ 1.2...... r ^{nr+l+r}
_ (n + 1) n (n ~ 1) ... (ri  r + 2) _ ^
1.2 r "^^ •
Ex. To prove that ,^1?^ = n^r + ^ • n^rV
A particular thing is ahsent in JP^ of the permutations of the
^n+ 1) things and occurs in ^^,^1; iSso when it does occur it can be
in eitner of r places. Hence
248. Theorem. To prove that, if x amd y he any two
positive integers such that oo + y = m, then will
Suppose that m letters a, 6,..., jp, },..., are divided
into two groups a, 6,..., and p, 3,..., there being x and
y letters respectively in these groups. Then the whole
number of sets of n out of the m things will clearly be
COMBINATIONS. 287
equal to the sum of the number of sets formed by taking
n out of the first group and none out of the second, n — 1
out of the first group and one out of the second, n — 2
out of the first group and 2 out of the second, and so on.
Now n can be chosen from the first group in ^C^ ways.
Also n — 1 can be chosen from the first group in ,(7^^
ways, and any one of these sets of n — 1 things can be
taken with any one of the ^C^ sets of 1 from the second
group, so ihat the numbfer of sets formed by taking w — 1
from the first group and 1 from the second is gC^__^ x ^(7^.
Similarly, the number of sets formed by taking n — 2
from the first group and 2 from the second is J^^^^ x ^C^.
And, in general, the number of sets formed by taking
n—r from the first group and r from the second is ^G^^ x ,(7^.
Hence we have
"t" • . . I z^n—r • w r < • • • • y^«*
If a? or y be less than n some of the terms on the right
will vanish; for ^C^ = if r > n.
249. Vandermonde's Theorem. From the last
Article, if x, y and n be any positive integers such that
a? 4 y is greater than w, we have, since JO^ = .— ,
r
\n \n
^n
1
Vi
+ 
n
^n^
y^
+ ...
n
I'
1^
.+
n
r
r
2*
*
r
2
+ ..
• •
.4^.
n
Multiply each side of the last equation by [w, and we
have
, . w (n — 1)
\n
r[ri — r
288 HOMOGENEOUS PRODUCTS.
The above has been proved on the supposition that x
and y are positive integers such that a? + y is greater than
n; and by means of the theorem of Art. 91, the proposi
tion can be proved to be true for aU values of w and y.
For the two expressions which are to be proved
identical are only of the nth degree in w and y. But, if
y has any particular integral value greater than n, the
equation is known to be true for any positive integral
value of w, and thus for more than n values; and
hence it must be true for that value of y and any value
whatever of x. Hence the proposition is true for any
particular value whatever of a?, and for more than n values
of y ; it must therefore be true for all values of x and for
all values of y.
This proves Vandermonde's theorem, namely : —
Ifnhe any positive integer , and x, y have any values
whatever; then will
/ \ n (n — 1)
r
r \n — r
Homogeneous Products.
250. The number of different products each of r
letters which can be made from n letters, when each
letter may be repeated any number of times, is denoted
by the symbol ^E^.
For example, the homogeneous products of two dimensions formed
by the three letters a, 6, c are a*, 6*, cr*, 6c, ca, db.
To find JI^.
Since in each of the rdimensional products of n
things there are r letters, the total number of letters in all
the products will be JI^ x r ; and, as each of the n letters
occurs the same number of times, it follows that the
HOMOGENEOUS PRODUCTS. 289
number of times any particular letter, a suppose, occurs is
^H^ X r ^ n.
Now consider all the terms which contain a at least
once. If any one of these terms be divided by a the
quotient will be of r — 1 dimensions ; and, when all
the terms which contain a are so divided, we shall obtain
without repetition all the possible homogeneous products
of the n letters of r — 1 dimensions. Now the homogeneous
products of r — 1 dimensions are in number „fl^i ; and,
by the above, the number of a's they contain is
r — 1
X ^H,^, Hence, taking into account the a which is
a factor of each of the „fl^i terms, the total number of a's
which occur in all the rdimensional products of the n letters
is
n^rl + ^ X n^rl, that IS „ff,.,.
Hence equating the two expressions for the number
of a*s, we have
n n
Since the above relation is true for all values of n and
r, we have in succession
n + r 2
r1
n^rl Z T X »r"r2>
rr __ n + r—S TT
n"2 "~ o X n"l'
7lfl
Also ^JETj is obviously equal to n.
S.A. 19
290 EXAMPLEa
Hence, by multiplying and cancelling common factors,
we have
H^ = "^"'^l.^g;;;;;^"^"^^ [See also Art. 293].
i»"^r
Ex. 1. Find the number of combinations three at a time of the letters
a, &, c, d when the letters may be repeated. Ans, 20.
Ex. 2. Find the number of different combinations six at a time which
can be formed from 6 a% 6 b% 6 c'b and 6 d'a. Am. 84.
Ex. 8. Shew that ^H,.=^_ifl^+^H^i,
and deduce that
Ex. 4. Shew that
251. Many theorems relating to permutations and
combinations are best proved by means of the binomial
theorem : examples will be founa in subsequent chapters.
[See Art. 292.]
Ex. 1. Find the number of ways in which mn different things can be
divided among n persons so that each may have m of them.
The number of ways in which the first set of m things can be
given is ,^C^ ; and, whatever set is given to the first, t£e second
set can be given in mnm^m w^Jb i ^o also, whatever sets are given
to the first and second, the third set can be given in fnnanfim ways;
and so on.
Hence the required number is
t
ntn^m ^ m(i»—l) ^m ^ m(n— 2) ^m ^ ^ TrnPrn ^ tnPm
\mn m(nl) \ m (n  2) (2m Im
\m m(n~l) \m m(n2) m \m (n  3) I^t? 'j^
jmn
Ex. 2. Prove that
Since ,>C,.J.=^ (^'^>^^^^+^> . ^(^+l)(^+^l)
* "^ * '^ [r (r
_ n^(ng~12) (narTia)
~ 12.2* r» •
EXAMPLES. 291
we haye to prove that
^ n* (n«  1«) n«(n«l«)(n8~2»)
lia+ 12.22 " ia.2^3» ■*"
n«(n'l«)...(n^nl^ _
+(1) 12.22. ..na "*
Now the first two terms = p— ;
thr^ = . <"'\l<g^'>
f„r.,  («' !')("' 2*) ("'8^
lour — — p 2'.3a *
and 80 on.
Henoe the sum of all the terms on the left
_ (n»l«)(na2») {n^n^) _^
~^~ ^ 12.22 n« """•
Ex. 3. Shew that n straight lines, no two of which are parallel and
no three of which meet in a point, diyide a plane into ^ (n+ 1) + 1
parts.
The nth straight line is cut by each of the other n 1 lines ; and
hence it is divided into n portions. Now there are two parts of the
plane on the two sides of each of these portions of the nth line which
would become one part if the nth line were away. Hence the plane is
divided by n lines into n more parts than it is divided by n  1 lines.
Hence, if F(x) be pnt for the number of parts into which the
plane is divided oy x straight lines, we have
F(n)=F(nl)+n.
Similarly
F(»l)=F(n2) + (nl),
and
F(2)=2'(l)+2.
But obviously
F{1)=2.
Henoe
F(n)=2+2+3 + 4+ +n
= 1 + Jn(n + 1).
Ex. 4. Suppose n things to be given in a certain order of succession.
Shew that the number of. ways of taking a set of three things out of
these, with the condition that no set shall contain any two things
which were originally contiguous to each other is ^ (n  2) (n  3) (n  4).
Shew also that if the n given things are arranged cyclically, so that
the nth is taken to be contiguous to the first, the number of sets is
reduced to n (n  4) (n  6).
19—2
292 EXAMPLES.
Consider the second case first.
Let the different things be represented by the letters a, 6, c,.....«
k, I.
Suppose that a is taken first. Then, if either of the two letters
next but one to a be taken second, any one of n  5 letters can be
taken for the third of the set. If, however, the second letter is not
next but one to a, but in either of the n5 other possible places,
there would be a choice of n  6 places for the third letter of the set.
Hence the total number of ways of taking 8 letters in order a being
first is 2(n5) + (n6)(n6'), that is (n4)(n6). There is the
same number when any one oi the other letters is taken first ; hence,
as the order in which &e three letters in a set are taken is indifferent,
tiie total number of sets is \n (n  4) (n ~ 5).
In order to obtain the first case from the second, we have only to
suppose that a and I are no longer contiguous. Hence the number
in tilie first case is the same as &at in the second with the addition
of those sets which contain a and 2, and there are n  4 of these.
Hence the number in the first case is
}n(n4)(n6) + (n4)=(n2)(n8)(n4).
Ex. 5. There are n letters and n directed envelopes : in how many
ways could all the letters be put into the wrong envelopes?
Let the letters be denoted by the letters a, 5, c... and the corre
sponding envelopes by a\ b\c\
Let F (n) be the required number of ways.
Then a can be put into any one of the n 1 envelopes &',(/,
Suppose a is put into kf ; then k may be put into a', in which case
there will be ^ (n  2) ways of putting all the others wrong. Also if
a is put into V, the number of ways of disposing of the letters so that
k is not put in a\ b not in b\ <&c. is ^(n~ 1).
Hence the number of ways of satisfying the conditions when a is
put into kf is F(nl)+F{n2), The same is true when a is put
into any other ot the envelopes V, c', .•• Hence we have
F(n) = (nl){F(nl)+^(n2)};
F(n)nF(nl)={F(nl)(nl)F(n2)}.
Similarly F(nl)(nl)F(n2)= {F(n2)(n2)F(n8)}
F (8)  SF (2) =  {F (2)  2F (1)}.
But obviously F (2) = 1 and F (1) = ;
F(n)nF(nl) = (ir.
„ Fhi) F(nl) , ,, 1
Hence r^ '  f — z^zsCl)". — .
EXAMPLES. 293
wxxix*«»xj nl n2 ^ ' jnl/
and ^?W^(l)al
Hence, by addition,
^<»)=fe{irE3^i+nr}
the number required.
EXAMPLES XXIIL
1. In how many different ways may twenty different things
be divided among five persons so that each may have four ?
2. A crew of an eightoar has to be chosen out of eleven
men, five of whom can row on the stroke side only, four on the
bowside only, and the remaining two on either side. How
many different selections can be made ?
3. There are three candidates for a certain office and twelve
electors. In how many different ways is it possible for them
all to vote ; and in how many of these ways will the votes be
equally divided between the candidates ?
4. Shew that ^JO^ : ^jG^ is equal to
1 .3.5 (4nl)
{1.3.5 (2wl)}''
5. Find the number of significant numbers which can be
formed by using any number of the digits 0, 1, 2, 3, 4, but
using each not more than once in each number.
6. Shew that in the permutations of n things r together,
the number of permutations in which p particular things occur
7. There are n points in a plane, no three of which are in
the same straight line ; find the number of straight lines formed
by joining them.
8. There are n points in a plane, of which no three are on
a straight line except m which are all on the same straight line.
Find the number of straight lines formed by joining the points.
294 EXAMPLES.
9. There are n points in a plane, of which no three are on
a straight line except m which are all on a straight Una Find
the number of triangles formed by joining the points.
10. Shew that the number of different nsided polygons
formed by n straight lines in a plane, no three of which meet
in a point, is ^ n  1.
11. There are n points in a plane which are joined in all
possible ways by indefinite straight lines, and no two of these
joining lines are parallel and no three of them meet in a point.
Find the number of points of intersection, exclusive of the n
given points.
12. Through each of the angular points of a triangle m
straight lines are drawn, and no two of the 3m lines are paiullel ;
also no three, one from each angular point, meet in a point.
Find the number of points of intersection.
13. The streets of a city are arranged like the lines of a
chessboard. There are m streets running north and south,
and n east and west. Find the number of ways in which a
man can travel from the N.W. corner to the S.E. comer, going
the shortest possible distance.
14. How many triangles are there whose angular points
are at the angular points of a given polygon of n sides but none
of whose sides are sides of the polygon ?
15. Shew that 2n persons may be seated at two round
tables, n persons being seated at each, in ^ different ways.
16. A parallelogram is cut by two sets of m lines parallel
to its sides : shew that the number of parallelograms thus
formed is J (m + l)'(m + 2)'.
17. Find the number of ways in which p positive signs
and n negative signs may be placed in a row so that no two
negative signs shall be together.
18. Shew that the number of ways of putting m things in
r»+l places, there being no restriction as to the number in each
place, is (m + n) !/m ! n 1
EXAMPLES. 295
19. Shew that 2n things can be divided into groups of n
2n
pairs in ^ ways.
20. Find the number of ways in which mn things can be
divided into m sets each of n things.
21. Shew that n planes through the centre of a sphere, no
three of which pass through the same diameter, will divide the
surface of the sphere into ti' — n + 2 parts.
22. Shew that the number of parts into which an infinite
plane is divided hj m^n straight lines, m of which pass
through one point and the remaining n through another, is
mn + 2m + 2w — 1, provided no two of the lines be parallel or
coincident
23. Find the number of parts into which a sphere is
divided hj mhn planes through its centre, m of which pass
through one diameter and the remaining n through another,
no plane passing through both these diameters.
24. Find the number of parts into which a sphere is
divided by a + 6 + c+... planes through the centre, a of the
planes passing through one given diameter, b through a second,
c through a third, and so on; and no plane passing through
more than one of these given diametera
25. Shew that n planes, no four of which meet in a point,
divide infinite space into ^ (n^ + 5w + 6) different regions.
26. Prove that if each of m points in one straight line be
joined to each of n points in another, by straight lines termin
ated by the points ; then, excluding the given points, the lines
will intersect Jmn (m  1) (w  1) times.
27. No four of n points lying in a plane are on the same
circle. Through every three of the points a circle is drawn,
and no three of the circles have a common point other than one
of the original n points. Shew that the circles intersect in
■,?5^(n 1) (ri2) (w3) (ri4) (2w 1) points besides the
original n points, assuming that every circle intersects every
other circle in two points.
CHAPTER XX.
The Binomial Theorem.
252. We have already [Art. 67] proved that the con
tinued product of any number of algebraical expressions is
the sum of all the partial products which can be obtained
by multiplying any term of the first, any term of the second,
any term of the third, &c.
253. Binomial Theorem. Suppose that we have n
factors each of which is a + 6.
If we take a letter from each of the factors of
(a + 6)(a + 6)(a+6)..."...
and multiply them all together, we shall obtain a term of
the continued product; and if we do this in every possible
way we shall obtain all the terras of the continued pro
duct [Art 67.]
Now we can take the letter a every time, and this can
be done in only one way; hence a* is a term of the
product.
The letter b can be taken once, and a the remaining
(n — 1) times, and the number of ways in which one b can
be taken is the number of ways of taking 1 out of n things,
so that the number is ^(7, : hence we have
BINOMIAL THEOREM. 297
Again ; the letter b can be taken twice, and a the
remaining {n — 2) times, and the number of ways in which
two 6*s can be taken is the number of ways of taking 2 out
of w things, so that the number is ^0,: hence we have
And, in general, b can be taken r times (where r is
any positive integer not greater than n) and a the re
maining n — r times, and the number of ways in which r
Vs can be taken is the number of ways of taking r out of
n things, so that the number is J)^: hence we have
n r
Thus (a+ b)(a + b)(a + b) to n factors
the last term being J)y^b\ i.e. 6*.
Hence, when n is any positive integer, we have
{a^by = Qr + J0^.ar'^b + jO^.oT^V + ...
The above formula is called the Binomial Theorem.
If we substitute the known values [see Art. 244] of
fp\> fP%> n^8»'" ^^ ^^^ series on the right, we obtain the
form in which the theorem is usually given, namely
1 • iU
...+7—7 ar^V+... + b\
\r [n — r
The series on the right is called the expansion of
(a + 6)*.
254. Proof by Induction. The Binomial Theorem
may also be proved by induction, as follows.
298 BINOBnAL THEOBEM.
We have to prove that, when n is any positive integer,
1. , JU
\r \n—r
or that ~
Now if we assume that the theorem is true when the
index is n, and multiply by another factor a + 6, we have,
when like terms of the product are collected,
Now l + ^(7, = l + n = „^,e,,
and, in general,
.a., + „a = ^fir [Art. 247].
Hence
Thus if the theorem be true for any value of n, it will
be true for the next greater value.
Now the theorem is obviously true when n = 1. Hence
it must be true when w = 2; ana being true when n= 2,
it must be true when w = 3; and so on indefinitely. The
theorem is therefore true for all positive integral values
of n,
Ex. 1. Expand (a + &)^.
We have
= o* + 4a86 + 6a262 + 4a63 + 54.
GENERAL TERM. 299
Ex. 2. Expand {2x  yf.
Put 2x for a, and  y for & in the general formula : then
= 8a:»  12a;*y + 6x^2  y'.
Ex. 3. Expand (a  &)^
Change the sign of & in the general formula; then we have
(ad)'»=a»+na**Mfe) + ^^r^«*"'(^P+
=o»no»>6+'!^^^o*»i»+
In
+(i)'"0^«*"^'"+ +(ir6.
255. General term. By the preceding articles we
see that any term of the expansion of (a + 6)" by the
Binomial Theorem will be found by giving a suitable
value to r in
In
Th — a^'i'
On this account the above is called the general term
of the series. It should be noticed that the term is the
(r + l)th term from the beginning. [See Note Art 244.]
256. Coefficients of terms equidistant respec
tively from the beginning and the end are equal.
In the expansion of (a + 6)" by the Binomial Theorem,
the (r + l)th term from the beginning and the (r + l)th
term from the end are respectively
^a, . a'^'V and ,0 . a*"6"^.
But «a = «C,.r. [Art. 245.]
300 GUEATEST TERM.
Hence, in the expansion of (a + 6)*, the coefficients of
any two terms equidistant respectively from ike beginning
ana the end are equal.
This result follows, however, at once from the £Ekct that
(a + 6)*, and therefore also its expansion, would be unaltered
by an interchange of the letters a and h ; and hence the co
efficient of oT*}/ must be equal to the coefficient of b*^a\
257. If, in the formula of Art. 263, we put a = 1 and
6 = a?, we have
(l+a?y = l + ^+ ; o V +... + 7— p= af +...+a?".
1.2 r p — r
This is the most simple form of the Binomial Theorem,
and the one which is generally employed.
The above form includes all possible cases: if, for
example, we want to find (a + by by means of it, we have
(a + 6)={a(l4)}''=a(l4^)"
.fi . 6 , n(nl)/b\*
= o" + na'b + H^UlD a»6» + ...
258. Greatest teim of a binomial expansion.
In the expansion of (1 + a?)*, the (r + l)th term is formed
from the rth by multiplying by w.
Now
n — r+l fn + 1 ^\ , n + 1
= f 1 J a?, and clearly
diminishes as r increases ; hence x diminishes
r
as r is increased. If os be less than 1 for any
GREATEST COEFFICIENT. 301
value of r, the (r + l)th term will be less than the rth.
In order therefore that the rth term of the expansion may
be the greatest we must have
n — r + 1 , jW — r— 1 + 1 ,
— a? < 1, and = a? > 1.
r r— 1
TT (n + l)a!  (n+l)a? , 
Hence r >  — ^—, and r < ^ ^ + 1.
The absolute values of the terms in the expansion of
(1 +xy will not be altered by changing the sign of a? ; and
hence the rth term of (1 — a?)* will also be greatest in
absolute magnitude if
(71+ 1) a? J ^ (n + l)a? 
r > ^ — — ^ , and r < ^^ 4— + 1
57+1 a?+l
If r = ^ —, then x= 1 ; and hence there
x + l r
is no one term which is the greatest, but the rth and
r+lth terms are equal, and these are greater than any of
the other terms.
Since (o+a;)*=a*f 1 +  j ,
the rth term of (a+x)** is the greatest when
(«+l)5 (n+l)^
f> and < +1.
?+l 2+1
a a
Ex. 1. Find the greatest term in the expansion of (l + x)^; when
1
21 21
The rth term is the greatest, if r> g and r<l+^. Hence
the fifth term is the greatest.
Ex. 2. Find the greatest term in the expansion of (1+a;)^^ when
5
The rth term is the greatest, if r > 5 and < 6. Thns there is no
one term which is the greatest, bat the 5th and 6th terms of the
302 EXAMPLES.
expansion are equal to one another and greater than any of the
other terms.
Ex. 8. Find the greatest term in the expansion of (lO+da;)^ when
07=4. Ans. The ninth term.
The greatest coefficient of a binomial expansion can
be found in a similar manner. For in the expansion of
(1 ± xy the coeflBcient of the (r + l)th term is formed from
that of the rth by multiplying by ± . Hence the
rth coe£Scient will be the greatest in absolute magnitude,
if < 1 and ; > 1.
r r — 1
That is if r>^?i and <H!^.
Hence when n is even, the coefficient of the rth term
is the greatest when r = ^ + 1 ; and when n is odd, the
coefficients of the— ^th and ^ th terms are equal
to one another and are greater than any of the other terms.
For example, in (1+a;)** the coefficient of the 11th term is the
greatest; and in (l + a;)" the coefficients of the 6th and 7th terms
are greater than any of the others.
EXAMPLES XXIV.
Write out the following expansions :
1. {x^af. 2. (2a jc/. 3. (la;^.
4. (2a3ay. 5. (2a^3)*. 6. (aj'2yy
7. Find the third term of {x  Zyf\
8. Fmd the fifth term of (3a;  4)~
9. Find the twentyfirst term of (2 — a;)".
10 Find the fortieth term of (a;  y)".
EXAMPLES. 303
11. Find the middle term of (1 + x)\
12. Find the middle terms of (1 + «)".
13. Find the general term of {x — 3y)*.
14. Find the general term of (aj* + y^y,
15. Write down the first three terms and the last three
termsof (3a;2y)".
16. Find the term of (l+as)*' which has the greatest
coefficient.
17. Find the two terms of (1 + a:)** which have the greatest
coefficients.
18. Shew that the coefficient of af in the expansion of
(1 + a?)*" is double the coefficient of x* in the expansion of
(1 + x)^'\
19. Shew that the middle term of (1 + xf is
1.3. 5... (2^.1) 2„^,
\n
20. Employ the binomial theorem to find 99*, 51* and
999^
21. Shew that the coefficient of a^ in the expansion of
IS
('"^^y^
22. Find the middle term oilx^ .
23. The coefficients of the 5th, 6th and 7th terms of the
expansion of (1 +a;)" are in arithmetical progression: find n.
24. For what value of n are the coefficients of the second,
third and fourth terms of the expansion of (1 + a:)* in arith<
metical progression %
25. If a be the sum of the odd terms and h the sum of the
even terms of the expansion of (1 + a;)", shew that
304 PBOPERTIES OF THE COEFFICIENTS.
259. ProperUes of the coefficients of a binomial
expansion.
It will be convenient to write the Binomial Theorem
in the form
(l+a?)* = Ca + Cia? + c^ + ... +cX+ ... c^a?* (i),
where, as we have seen, c^ = c„ = 1 ; c^ = c^.j = n;
and. in general. c^ = c.^ = J^.
I. Put a? == 1 in (i) ; then
2'' = Co + Ci + Cj+... + c^.
Thus the sfum of the coefficients in the expansion of
(l\xyisT.
II. Put a?=s — 1 in (i) ; then
(ll)" = Coc,+c. +(irc.;
.. = (Co + Cj + c^ + . . . )  (Cj + C3 + Cj + . . . ).
Thus the sum of the coefficients of the odd terms of a
binomial expansion is equal to the sum of the coeffi^ents
of the even terms,
m. Since c^ = c^.^, we have
(l+xy = c^ + c^x+c^ + ... +cX+ + c^^%
and (1 +xy — c^ + c^^yX + c^.^ + ... c^.^ + ... +0^.
The coefficient of af^ in the product of the two series
on the right is equal to
Co+c^* + c^^+ +c^\
Hence [Art. 91] c* + c,"+ ... +c; + + c/
is equal to the coefficient of a?* in (1 + xy x (1 +a?)*, that
2n
is in (1 +x)^: and this coefficient is ^^= •
'^ n n
PROPERTIES OF THE COEFFICIENTS. 306
Hence the swnt of the squares of the coefl&cients in the
2n
expansion of (1 + x)* is 7=^ •
^ \n\n
IV. As in III, we have
(1 + a?)" = Co + CjO? + c^a?' + ... + c^a?**,
and (1  xY = c^  c^.^a? + c^.^' +... + ( l)"cX.
The coeflScient of af in the product of the two series
on the right is equal to
The coefl&cient of af in (1+ ocj' x (1 — a?)*, that is in
 1^
(1 — a?y, is zero if n be odd, and is equal to (— l)^ rTj—
if n be even.
Hence c^  c^* + c,'  . . . + ( 1)X* is zero
n
or (— l)*iil/(in!)*, according as w is odd or even.
Ex. 1. Shew that
Ci + 2c2+ 3cj+ ... +rc,.+ ... +?M;,j=n2*i.
WehaTO Ci + 2c3+3c3+...+nc^
''"*'^i:2*^ 1.2.3 + + ^7F^r++^
n1
=Ji+(„_i)+(zLzi)(^)+ +4t^+ +4
I ^ ' 1.2 rlnr j
=n(l + l)*i=n2»i.
11 c 1
Ex.2. Shew that CosCi + o<^3 + (1)*^^ = — n*
"2*3'' ' w+1 n+1
XTT I. 1 1 •. ^ 1 In(nl) .
We have Co2Ci+gCa&c.=l^n+^ — ^^g— '&C.
S.A. 20
306 PBOPEBTiES OF THE cx>BrnciEin&
"i+1 i"^* Ti^ 1.2.3 — +<*'^r
«+l « + l I 1.2 1.2.3 I
= ?. i (ll)»+>=^ .
«+l fl+1* ' «+l
Ex. 3. Shew that, if s be any poaitire integer.
X x+1 x + 2 ^* '^x+B x(x + l)...(x+«)
iltfiMM that
X x + l^x+2 ^^ ' x+fi x(x+l)...(x+i»)*
for oU voliief of x, and for any particular valae of ».
Change X into x+1; then
x+1 x+2^x + 3 ^* ' x+n+1
H
(x+1) (x + 2) ... (x+n+ 1) *
HencOf by subtraction,
x" X + 1 "^ X+2 ■*"^" ' x+r "^
1 ,n + l
+ (.1)*+! — I =s ' .
^^ ' x+n+1 x(x+l) (x+n + 1)
Bot n^r+ffiri=%^i^rf ^^^ ^ values of r [Art. 247].
Hence we have
X x+1 x + 2 ^ ' x+n+1 X (x + 1) ...(x+n+1)*
Hence if the theorem be tnie for any particular value of n it will
be true for the next greater value. But the theorem is obviously true
for all values of x when n=l: it is therefore true for all positive
integral values of n. [See also Art. 297, Ex. 3.]
By giving particular values to x we obtain relations between
CofCif Ao, For example :
Put X = 1 ; then we have
1 ~ 2"^3 " "rT+l' I
PKOPERTIES OF THE COEFFICIENTS. 307
Put x =  ; then f  ^ + f 
2' 1 3 ' 6 1.3.5... (2» + l)"
Ex. 4. Shew that
CoaCi(al)+C2(a2)C8(a3) + + (l)'»c„(an)=0,
and that
CoaaCi(al)3 + c2(a2)2.c8(a3)3+ + (!)» c„ (a n)a=0.
We have from II., if n be any poBitive integer,
Hence, if n > 1
l(nl)+ '"y',"^'  + (l)=0 (n).
Multiply (i) by a and (ii) by n and add ; then
an(aX) + ^^^(a2) + (!)'» (an)=0 (iii),
where n is > 1.
Change a into a1 and n into n  1 in (iii) ; then, n being > 2, we
have
al(nl)(a2) + +(l)»»i(an)=0 (iv).
Now multiply (iii) by a and (iv) by n and add; then
aa»(al)2+!?i^:ifJ(a2)2 + (l)'»(an)3=0,
provided n is greater than 2.
By proceeding in this way we may prove that
aPn(al)*+^^J^^(a2)P + (l)»*(an)P=0,
provided that p is any positive integer less than n.
[See also Art. 305.]
260. Continued product of n binomial factors of
the form xha, x + b, x + c, &c.
It will be convenient to use the following notation :
Sj is written for a+6Hc+..., the sum of all the
letters taken one at a time. 8^ is written for oft + oc +.. .,
the sum of all the products which can be obtained by
20—2
308 PRODUCT OF BINOMIAL FACTORS.
taJdng the letters two at a time. And, in general, S^ is
written for the sum of all the products which can be
obtained by taking the letters r at a time.
Now, if we take a letter from each of the binomial
factors of
(a?Ha)(a? + 6)(a? + c)(a?f d) ,
and multiply them all together, we shall obtain a term
of the continued product ; and, if we do this in every
possible way, we shall obtain all the terms of the con
tinued product.
We can take x every time, and this can be done in
only one way; hence i^ is a term of the continued
product.
We can take any one of the letters a, 6, c..., and x
from all the remaining n—l binomial factors ; we thus
have the terms cw?**"', 6df"\ caf "^ &c., and on the whole
8^ . ar'\
Again, we can take any two of the letters a, 6, c...,
and X from all the remaining n — 2 binomial factors;
we thus have the terms abx , aca*''\ &c., and on the
whole /Sj . af'\
And, in general, we can take any r of the letters
a, 6, c..., and x from all the remaining n — r binomial
factors ; and we thus have 8^ . a?""*".
Hence (a?+a)(a? + 6)(a? + c)
the last term being a6cd , the product of all the
letters a, 6, c, d, &c.
By changing the signs of a, 6, c, &c., the signs of
Sj, Sg, Sj, &c. will be changed, but the signs of /S^, 8^, 8^y
&c. will be unaltered.
Hence we have
(a? — a)(aj — 6)(a; — c)
=x^8,.x'''+8,,x'...+ (iyS,.ar... + (^iyabcd...
VANDBRMONDE*S THEOREM. 309
261. Vandermonde's Theorem. The following
proof of Vandermonde's Theorem is due to Professor
Cayley*. [See also Art. 249.]
We have to prove that if n be any positive integer,
and a and b have any values whatever ; then will
(a + 6X = ^n + '^n A + \ 2 ^«*« "^ •••
Assume the theorem to be true for any particular value
of n. Multiply the left side by a + 6 — n ; it will then become
(a + 6)n+i. Multiply the successive terms of the series on
the right also by a + 6 — w but arranged as follows : —
for the first term {{a — ti) + 6} ;
for the second {(a — w + 1) + (6  1)} ;
and for the rth {(a w + r l) + (6 — r + 1)}.
We shall then have
+ nC;.an.2i^,{(a^ + 2) + (62)} + ...
+nOr^.a,,^,b,.^{(an + r'l) + (br + l)}
+ ... +6„{a + (6n)}.
Now a^ {(a w) + 6} = a^i + a^ 6^,
,0,_i.a,^,6^i{(an + rl) + (6r + l)}
•^r • <^nrK {(a  W + r) + (6  r)} = JJr Krfl^r + ^nr^r+l)
* Messenger of MathematicSt Vol. v.
810 MULTINOMIAL THEOREM.
Hence (a + 6)^^ = a„+i + (1 + J}^ aj>^ + . . .
= a„4.1 + „+iOj anfej + . . . + „+iC^r ^n+lr&r + • • • + ^n+l*
since JJ^^ + «(7^ = ^lO^ .
Thus, if the theorem be true for any particular value
of n, it will also be true for the next greater value. But
it is obviously true when w = 1 ; it must therefore be true
when w = 2 ; and so on indefinitely. Thus the theorem is
true for all positive integral values of w.
262. The Multinomial Theorem. The expansion of
the Tith power of the multinomial expression a + 6 + o + ...
can be found by means of the Binomial Theorem.
For the general term in the expansion of
(a + 6 + c + d + . ..)'*, that is of {a +(6 + c + d+ ...)}%
by the Binomial Theorem is
\n
rr= — a** (6 + c + d + ...)*'•■.
Similarly the general term in the expansion of
(6 + c + d+.. .)•*•■
by the Binomial Theorem is
\n — r
The general term in the expansion of (c + d + . . .)*"**"' by
the Binomial Theorem is
8
J —(f{d+ ...)""^*
t \n — r~8 — t
Hence the general term in the expansion of
{a^h + c + d^ ...y
\n — r — 8 ^ ^
, , , » t * d c ••••
\t n — r — s—t
MULTINOMIAL THEOREM. 311
\n
that is , , 'tx — a*'6'c'...,
where each of r, «, ^ ... is zero or a positive integer, and
The above result can however be at once obtained by
the method of Art. 253, as follows.
We know [Art. 67] that the continued product
(a + 6 + c+ ...)(a + 6 + c ...)(a + 6 + c + ...)...
is the isum of all the different partial products which can
be obtained by multiplying any term from the first multi
nomial factor, any term from the second, any term from
the third, &c.
The term a*" 6* c* . . . will therefore be obtained by taking
a from any r of the n factors, which can be done in jO,
different ways; then taking b from any s of the remaining
n~r factors, which can be done in ,^0, different ways ;
then taking c from any t of the remaining n — r — s factors^
which can be done in ^^^fit different ways; and so on.
Hence the total number of wa^s in which the term
a*" 6' &... will be obtained, which is the coefficient of the
term in the required expansion, must be
that is
In In — r In — r— « In
Us X — ' — X — ' X = *—
\r n — r «nr— « [^n— r— 5 — < *" jrls^...*
Hence the general term in the expansion of (a + 6 + c + . . .)"
is
n
. • .
312 EXAMPLES.
Ex. 1. Find the coefficient of abc in the expansion of (a + & + c)*.
The required coefficient = r" = 6.
Ex. 2. Find the coefficients of a^b^, bcd^ and ahcd in the expansion
of (a+6 + c + d)*.
We have the terms
•
Thns the required coefficients are 6, 12 and 24 respectively.
263. By the previous Article, the general term of the
expansion of {a + bxica^ + da? + )* is
\n
that IS , , 7, dl'Vc^d!' a?*^****^— .
Hence to find the coefficient of any particular power of a?,
say of af , in the expansion, we must find all the diflferent
sets of positive integral values of r, 5, ^,... which satisfy
the equations
« + 2* + 3«A+ *=a,
r + 5 + ^ + w+ =n.
The required coefficient will then be the sum of the
coefficients corresponding to each set of values.
Ex. 1. Find the coefficient of a^ in the expansion of (1 + 2d; + Sac*)^.
The general term is . ~.^ 2^«a!*+*, and the terms required are
till!
those for which «+2t=5 and r+«+t=4.
Since each of the quantities r, s and t must be zero or a positive
integer, the only possible sets of vtJues are t=2, «sl, r=l and
t=l, «=3, rs=0, the corresponding coefficients being ~ .^^ .2.3^
II liilH
/
EXAMPLES. 313
14
and  .2^.3, that is 216 and 96 respectively. Hence the
12 1? II
required coefficient is 812.
In simple oases the resnlt can be readily obtained by actual
expansion. We have
(l + 2aj+3a?y=l + 4(2aj + 3a!?) + 6(2aj+3a^)>+4(2a:+3a;9)«+(2a; + 3a;9)*.
Only the last two terms will contain x^ and the coefficients of o;*^ in
tiiese terms wiU be found to be 216 and 96 respectively, so that the
reqidred coefficient is 312.
Ex. 2. Find the coefficient of x^ in the expansion of (1 + a; + a^)\
Afu, 6.
Ex. 3. Find the coefficient of o^ in the expansion of (1 + ^ + as?)!
Ans. 16.
Ex. 4. Find the coefficient of x^ in the expansion of (2 +x  d^)!
Afu, 0.
Ex. 5. Find the coefficient of x^^ in the expansion of
(7+a;+a;2+ic"+aj*+x*)«. Aru. 39.
Ex. 6. Find the coefficient of the middle term of the expansion of
(1 + a: + a;2 + a;» +«*)«. Atu. 381.
EXAMPLES XXV.
1. Prove that
c„2c, + 3c, +(l)"(n + l)c, = 0.
2. Prove that
c,2c, + 3c3 +(l)""'wc, = 0.
3. Prove that
c^+2cj + 3c,+ + (» + l)c, = 2"'"'(w + 2).
4. Prove that
c, + 2c3 + 3c^ + +(wl)c,= l + (w2)2""\
5. Prove that
c^ + 3c, + 6c,+ + (2w + l)c„ = (n+l)2".
6. Prove that
3c, + 7c,+ llc3+ + (4wl)c,= l+(2nl)2".
314
7. Prove that
e, c. c, c 2*»l
12 3 n + 1 w+1 •
8. Prove that
C^ C C. C 9."
:r +;? +^ += +
13 5 7 w+1*
9. Prove that
^1 <^3 <?6 21
2 ^4 ^6^ n\l*
10. Prove that
c. c, « c l+n 2"**
s + if +f + +
2 3 "^4 w + 2 (7Hl)(n+2)*
11. Prove that
Cj c, C3 ,  J c^ __ 1 1 1
23 ^'wl2 n
12. Prove that
1 4 7 ^^ ^ 371+1 ~ 1.4. 7. ..(371 + 1)'
13. Prove that
\2n
<^0^ + ^,^r+, + +<^n.rC.=
n + r\n — r
14. Prove that, if
(l+xy = c^ + CjX + c^ + iCJX^y
then 7i(l+aj)"*=c, + 2Cgaj + 3c3aj*+ +wc,a5""*,
and {l + (7i+l)a;}(l +a;)"* = Co + 2c,a; + + (w+l)cX
Hence prove that,
. ^ . I2n1
c,» + 2c," + 3C3" + +wc/ =
n~l hr
, « ^ o « . (»+2)2wl
and c/ + 2c,V3c/ + +(n+l)cj= ^ , /^— ; —
EXAMPLES. 315
15. Shew, by expanding {(1 +«)" I}"*, where m and n
are positive integers, that
16. Prove that, if ti > 3,
(i) aw(al) + ^^$^i^(a2). + ( l)(aw) = 0.
(ii) a5w<al)(6l) + '?^^^^^(a2)(62)
' + (l)"(aw)(6w) = 0.
(iii) a5cw(al)(6l)(cl)+^^^^(a2)(62)(c2)
+ (l)(aw)(6n)(cw)=0.
17. Shew that, if there be a middle term in a binomial
expansion, its coefficient will be even.
18. Shew that the coefficient of x" in the nth power of
as* + (a + 6) 05 + aft is
19. If w be a positive integer and P^ denote the product of
all the coefficients in the expansion of (1 + oj)", shew that
P. l« ■
20. Shew that
{lx)' = {l +x)'2nx{l +x)''+^^^^^!^ 0^(1 +x)'"
21. Shew that, if ?i be a positive integer,
 1+05 nin—l) l + 2o5
1+nx 1.2 (l+Tia:)'
_ rt(ytl)(w2) l+3 a; _^
1.2.3 (l+ni)""*" " •
22. Shew that
(a + b + c + d + ey = la' + 52a*6 + lOSa'6* + 202a'6c
+ 30Sa'6"c + 60Sa"6cc^ + 1 20abode.
316 EXAMPLES.
23. If {l+x + xy = aQ + ajX'{ajx^+ ,
n(nl) (iriiL ^
prove that a, wa,_, + —jy—'a,_. + __a^ = 0,
unless r is a multiple of 3.
24. Shew that, in the expansion of (1 + oj + 05*+ + re")",
where n is a positive integer, the coefficients of terms equi
distant from the beginning and the end are equal.
25. If %, Aj, a^, be the coefficients in the expansion of
{1+X + a?Y in ascending powers of ic, prove that
«.'<+«.• + ( ir'«..'=iK(i)"0
26. If (1 + a; + sc")" = a^ + a^x + a^aj' + a^x^ + ...,
prove that
a^a^a^a. + a.a, =0.
27. Shew that, in the expansion of (a^ + a, + a, + . . . + a^)*,
where n is a whole number less than r, the coefficient of any
term in which none of the quantities a^ a,, <bc. appears more
than once is n\
28. Shew that, if the quantities (1 + oj), (1 + 05 + 05'), ,
(1 + 05 + 05* + +05") be multiplied together, the coefficients
of terms equidistant from the beginning and end will be equal;
and that the sum of all the odd coefficients will be equal to the
sum of all the even, each being ^ (71 + 1) 1
29. Shew that the coefficient of 05" in the expansion of
(1 +aj+fl5*)" is
ninl) n{nl){n2){n^)
n(nl)(n2)(^3)(n~4)(n5) .
■*■ 1*.2*.3*
30. Shew that 18 can be made up of 8 odd numbers in
792 different ways, where repetitions are allowed and the order
of addition is taken into account.
CHAPTER XXL
CONVERGENCY AND DIVERGENCY OF SERIES.
264.. A series is a succession of quantities which are
formed in order according to some definite law. When a
series terminates after a certain number of terms it is said
to be 2,fimte series, and when there is an endless succession
of terms the series is said to be injmite.
We have already found that when the common ratio of
a geometrical progression is numerically less than unity
the sum of n terms will not increase indefinitely, but that
the sum will become more and more nearly equal to a
fixed finite quantity as w is increased without limit. Thus
the sum of an infinite series is not in all cases infinitely
great.
When the sum of the first n terms of a series tends to
a finite limit ^8^, so that the sum can, by sufficiently
increasing n, be made to diflfer from 8 by less than any
assignable quantity, however small, the series is said to b^
conver^ren^, and fif is called its fiwm. Thus l + i + J + ^ + ...
is a convergent series whose sum is 2.
When the sum of the first n terms of a series increases
numerically without limit as w is increased indefinitely, the
series is said to be divergent. Thus 1 + 2 + 3+4+.. . is a
divergent series.
318 CX)NVERGENCY AND DIVERGENCY.
When the sum of the n first terms of a series does not
increase indefinitely as n is increased without limit, and yet
does not approach to any determinate limit, the series is
neither convergent nor divergent. Such a series is some
times called an indeterminate or a neutrai series*.
For example, the series 1 — 1 + 1 — l + ...isan indeter
minate series, for the sum of n terms is 1 or according as
n is odd or even.
It is clear that a series whose terms are all of the same
sign cannot be indeterminate, but must either be conver
gent or divergent. For unless the sum of n terms increases
without limit as n is increased without limit, there must
be some finite limit which the sum can never exceed, but
to which it approaches indefinitely near.
265. If each term of a series be finite, and all the terms
have the same sigu, the series must be divergent. For, if
each term be not less than a, the sum of n terms will be
not less than noL and na can be made greater than any
finite quantity, however large, by sufficie^ increasing rl
266. The successive terms of a series will be denoted
by t^j, w,, Wj,. . . ; and, since it is impossible to write down all
the terms of an infinite series, it is necessary to know how
to express the general term, u^, in terms of n.
The sum of the n first terms will be denoted by U„ ;
and the sum of the whole series, supposed convergent, in
which case alone it has a sum, will be denoted by u.
Thus U=U^ + U^ + U^+ ... +t^n + ^iHl + *»
and ^n = ^i + ^8 + ^8+ ••• + ^n
267. In order that the series u^, u^, t^,, u^, ,u^,
^1.41 » ^' ^*y ^^ convergent it is by definition necessary
and sufficient that the sum
* These series are howeVer caUed divergent series by Caachy, Bertrand,
Laurent and others.
CONVERaENCY AND DIVERGENCY. 319
should converge indefinitely to some finite limit ^ as 71 is
indefinitely increased.
Hence Z7^, U^+^y U^^, &c. ... must differ from U, and
therefore from one another, by quantities which diminish
indefinitely as n is increased without limit.
Now ^n+l ^n = ^«+l>
Hence, in order that a series may be convergent, the
(n + l)th term must decrease indefinitely as n is increased
indefinitely, and also the sum of any number of terms
beginning at the (n + l)th must become less than any
assignable quantity, however small, when n is indefinitely
increased.
For example, the series 7 + ^ + 7+... + +... oannot be oon
12 3 n
yergent, although the nth tenn dimmlshes indefinitely as n is increased
indefinitely; for the snm of n terms beginning at the (n+l)th is
— r+ — rs+ — + zri which is greater than Trxn, that is, greater
n+1 n+2 2n 2n
than ^.
268. We shall for the present consider series in which
all the terms have the same sign ; and as it is clear that
the convergency or divergency of such a series does not
depend on whether the signs are all positive or all negative,
we shall consider all the signs to be positive.
The convergency or divergency of series can generally
be determined by means of the following theorems.
269. Theorem I. A series is convergent if all its
terms are less than the corresponding terms of a second
series which is known to he convergent.
320 GONYEBGENCT AND DIYERGENCT.
Let the two series be respectively
U=u^ + u^ + u^+
and ^—^i Ht', + tV+
Then, since u^<v, for all values of r, it follows that U
is less than V. Hence, as F is finite, U must also be
finite: this proves the theorem, for a series must be
convergent when its sum is finite and all the terms have
the same sign.
It can be proved in a similar manner that a series
is divergent if all its terms are greater than the
corresponding terms of a divergent series.
Ex. (i). To Bhew that the series = + r^ + =— ^^ + 1—55^ + — ^
conveigeDt.
The terms of the series are less than the terms of the series
j + j^+ ^ 2,2 "*" 1.2.2.2 "*""•* ^^ *^ ^^^ ^"^ " * ^^'
metrical progression whose common ratio is , which is therefore
known to he a conyergent series. The given series must therefore
also be convergent.
Ex. (ii). Shew that the series
(a+x) {a+x){2a+x) (a fa?) (2a + a;) (3a + x)
{b+x) {b\x){2b+x) (6+a;)(26 + a;)(36+x)
is convergent if a, 6 and x are aU positive, and a < &.
The terms of the given series are less than the corresponding
. M.^ a+x (a + x)* (a+x)^
terms of the series — ■ — 4 ^ —  — ^ +  — ■ — — +
[since , < r— if r > 1, a, 6 and x being positive and 6 > a].
The latter series is convergent, and therefore also the given series.
To ensure the convergency of the first series it is not
necessary that all its terms should be less than the
corresponding terms of the second series, it will be
sufficient if all the terms except a, finite number of them
CONVERGENCY AND DIVERGENCY. 321
be less than the corresponding terms of the second, for the
sum of a finite number of terms of any series must be
finite.
4 43 4? 44 4<^ 4<<
Ex. Shew that the series ^+12 + 13 +74 + 15 + ig +l7"*"— ^ ^^^'
vergent.
From the sixth term onwards, each term is less than the corre
45 46
sponding term of the series ftc + Halt + •••• Hence the series
beginniDg at the sixth term is convergent, and therefore the whole
series is convergent.
270. Theorem Hi If the ratio of the corresponding
terms of two series be always finite, the series mil both be
convergent or both divergent
Let the series be respectively
U=u^ + u^ + u^+ ,
and F = Vj + Vj +v, +
Then, since the quantities are all positive, ^ must lie
between the greatest and least of the fractions — [Art. 113].
Vf,
Hence CT : F is finite. It therefore follows that if U is
finite 80 also is V, and if t/* is infinite so also is V.
For example, the two series ^ + i5^ + + _l_+...
and T + + + + ... are hoth convergent or both divergent.
8r 1
For the ratio of the rth terms, namely . — r^ 7 :  is equal to
•^ (r + l)(r + 2) r
; — rr; — T , which is > 1 and < 8 for all values of r. Now we have
(r+l)(r+2)*
already proved that the second series is divergent: the first series is
therefore also divergent.
271. Theorem III. A series is convergent if after
any particular term, the ratio of each term to the preceding
is always less than some fixed quantity which is itself less
than v/nity.
S. A. 21
822 CONVERGENCY AND DIVERGENCY.
Let the ratio of each term after the r'** to the preceding
term be less than k, where ifc < 1.
Then, since ^<ifc, ^»<^^ ,
we have
Wr + «*r+i + ^r+a + < U^(l + k + I(^ \ )
< = — ^ , since k is less than 1.
l — k
Hence the sum of the series beginning at the r^ term
is finite, and the sum of any finite number of terms is
finite ; therefore the whole series must be convergent.
272. Theorem IV. A series is divergent if, after
any particular temriy the ratio of each term to the preceding
is either equal to v/nity or greater than unity.
First, let all the terms after the r^ be equal to m^;
then Wr+i + ^r+s + ••• + ^ii+r = w^r> ^^^ ^^r cau bo made
greater than any finite quantity by sufficiently increasing
n. The series must therefore be divergent.
Next, let the ratio of each term, after the r*^ to the
preceding term be greater than 1.
Then u^^^ > u^, u^ > w^^, > u^, &c.
Hence u^^,^ + u^^ + ... + u^^ > nu/, the series must
therefore be divergent.
1 2 22 2* 2*!
Ex. 1. In the series T+« + ";r + T + + + » tl^® ratio
12 3 4 n
^^ = = , which is greater than 1 ; the series is therefore
u^ n+1
divergent.
Ex» 2. In the series l^\2^x+S^x^+ , the test ratio is ^^\^^ x,
that is ( 1 H — j X. Now, if a; be less than 1, and any fixed quantity k
be chosen between x and 1, the test ratio will be less than k for all
terms after the first which makes
(l + l)V*.V*.te.n.^^.
CONVERGfiNCY AND DIVERGENCY. 823
fienoe the series is conyergent if ^ < 1.
If «= 1 the series is 1* + 2' + 8^ + whioh is obyionsly divergent,
and if a;> 1 the series is greater than 1^+2^+8'+
Thns the series l^+^x + B^a:^+ is divergent except when x
is less than onity.
273. When a series is such that after a finite number
of terms the ratio ^^ is always less than unity but
becomes indefinitely nearly equal to unity as ti is in
definitely increased, the test contained in Theorem III.
fails to give any result ; and in this case, which is a
very common one, it is often difficult to determine whether
a series is convergent or divergent.
For example, in the series
1»"^2*'*"3*"^4*"^
the ratio
^«*i _ ^*
u^ {n+iy
7 — T7'
Hence, if ifc be positive, the test ratio is less than
unity, but becomes more and more nearly equal to unity
as 71 is increased.
We cannot therefore determine from Theorem III.
whether the series in question is convergent or divergent.
Ill
274. To shew that the series t» f „» + q* + ••• ^ c^*"
vergent when k is greater than unity, and is divergent when
k is equal to unity or less than unity.
First, let k be greater than unity.
Since each term of the series is less than the pre
ceding term, we have the following relations:
2* "*■ 3* ^ 2* '
21—2
324 CONVERGBNCY AND DIVERGENCY.
4* ■*■ 5* "^ 6* ■*■ 7* ^ 4* '
aiid 2"* "*" (2* + 1)* "^ "^ (2""*"*  1)* ^ 2"* '
Hence the ^hole series is less than
I » "T* 2* "^ 4j» "^ Q* "^ •••••• "T" ank "T" •••>
that is, less than
l,_±_ 1 1 _1_
■1 •" 2*i "^ 2*^*"^* 2'^*"^* 9»«(*i) T" ••• •
But this latter series is a geometrical progression
whose common ratio, ^jj , is less than unity, since i > 1.
Hence the given series is convergent.
Next, let & = 1; then we can group the series as
follows :
1 1 . ri 11 ri 1 1 1"1
r 1 1 1^1
L2»' + 1 "^ 2*' + 2 "^ "^ 2*J "^
therefore, as each group of terms in brackets is greater
than 4» ^^® given series taken to 2* terms is greater
than l+4 + i + 4 + taken to n + 1 terms, that is,
greater than 1 + ^w, which increases indefinitely with n.
Hence 7 + 5 + + ^ divergent
Lastly, let k be less than unity; then each term of the
series rx + o* + ^ greater than the corresponding
term of the divergent seriesY + o + > ^^® series is
X iU
therefore divergent when k<l.
CONVERGBNCY AND DIVERGENCY. 825
276. The convergency or divergency of many series
can be determined by means of Theorems I. and IL, using
the series of the last Article as a standard series. The
method will be seen from the following examples.
Ex. 1. Is the series whose general term is ^ — r convergent or
diyergent ?
Sinoe^>i,if »>l,itfoUowsthat2^>2i. Butzi
2n
is diyergent; therefore 2 »tt ^ ^'^ diyergent
Ex. 2. Is the series whose general term is ^ — = convergent or
divergent?
„ n+2 n+2 8»i 3 __ « »+2 ._ 1 t, . « 1
^''^i^Tl^^r^v^^^' Hence 2^jj^<3S,. ButS,
is convergent [Art. 274]; therefore 2 , ^ is also convergent.
276. We have hitherto supposed that the terms of the
series whose convergency or divergency was to be deter
mined were all of the same sign. When, however, some
terms are positive and others negative, we first see whether
the series which would be obtained by making all the
signs positive is convergent; and, if this is the case, it
follows that the given series is also convergent ; for a con
vergent series, all of whose terms are positive, would
clearly remain convergent when the signs of some of its
terms were changed. If, however, the series obtained by
making all the signs positive is a divergent series it does
not necessarily follow that the given series is divergent.
For example, it will be proved in the next Article that
the series  — 4 + J— i+... is convergent, although the
series  + i + J+i + ... is divergent.
277. Many series whose terms are alternately positive
and negative are at once seen to be convergent by means
of
826 OONVERGBNCY AND DIVERGENCY.
Theorem V. A series is convergent when its terms
are alternately positive and negative, provided ea>ck term is
less than the preceding, and that the terms decrease without
limit in ahsohtte magnitvde.
Let the series be
^1 "" '^8 + ^8 "" ^4 + • • • ± ^1. ^ ^•+1 ± ^i.+« ^
By writlDg the series in the forms
^i^» + (^8^4) + K^«) + •••>
and ^iK'«^8)K^6) >
we see that, since each term is less than the preceding, the
sum of the series must be intermediate io u^^u^ and u. ;
and hence the sum of the series infinite. It is also similarly
clear that the absolute value of U—U^ is intermediate to
the absolute values of t^»+, — t^n+g and t*^^,, and therefore
U—TJ^ becomes indefinitely small when n is increased
without limit. The series must therefore be convergent.
For example, the series t~o + o~7 + "• convergent, since
the terms are alternately positive and negative and decrease without
limit. The series T~n + q~T + ^ ^^^ however a convergent
series although its sum is a finite quantity between  and 2, for the
?i + 1
nth term, namely , does not diminish indefinitely as n is
indefinitely increased.
278. We will now apply the preceding tests of con
vergency to three series of very great importance.
I. The Binomial Series. In the binomial series,
namely
, m(ml)... ( m/i + l) ^, .
CONVERGENCY AND DIVERGENCY. 327
the number of terms is finite when m is a positive
integer; but when m is not a positive integer no one
of the factors m, m — 1, m — 2, &,c. can be zero, and
hence the series must be endless.
To determine the convergency of the series when m
is not a positive integer we must consider the ratio
^«t+i • ^w •
Now i?**» = ^i:i^^.: = a,fl^i±l).
u^ n \ n J
Hence, for all values of n greater than m + 1, w„+i, and
u^ have different signs when x is positive, and have the
same sign when x is negative. Moreover, as n is in
creased, the absolute value of u^Ju^ becomes more and
more nearly equal to x. If therefore x be numerically
less than unity, the ratio u^+Ju^ will, either from the
beginning, or after a finite number of terms, be numeri
cally less than unity. Hence by Art. 271 the series
formed by adding the absolute values of the successive
terms will be convergent, and therefore also the series
itself must be convergent, whether its terms have all
the same sign or are alternately positive and negative.
Thus the binomial series is convergent, if x is numeri
cally less than unity*.
IL The Exponential Series. In the exponential
series, namely
l+a? + 2 + 3 + + [^ + »
the ratio u^^,Ju is x/n. Hence the ratio u Ju^ is nu
merically less than unity for all terms after the first for
which n is numerically greater than x. The series is
therefore convergent for all values of x.
* The series is also oonTergent when « = 1, provided n >  1 ; and it is
convergent when x=?:  1, provided n > 0, (See Art. 338.]
328 CONYEBGENCT AND DIYERQENCY.
IIL The iKigaritlunic Series. In the logarithmic
series, namely
^"2^3" "(^) n + ' '
the ratio u^Ju^ is = = — a? [1 = ) ; and hence
"+^* n + 1 \ w + 1/
u^Ju^ will be numerically less than unity provided x
is numerically less than unity. The logarithmic series
is therefore convergent when x has any value between
 1 and + 1.
If ^7=1, the series becomes 1 —^H^—..., which is
convergent by Theorem V.
If aj = — 1, the series becomes — (1+4 + J4 ...)> which
is known to be divergent. [Art. 274.]
279. The condition for the convergency of the product
of an infinite number of factors, and also some other
theorems in convergency, will be proved in a subsequent
chapter, [See Art. 337.] The two important theorems
which follow cannot however be deferred.
280. If the two series
and F=i;o + Vj^ + Vsa?* + +^^5?" + ...,
be both convergent, and the third series
P = t«oVo+ K^, + ujo^X'\{ujo^'¥u^v^ + ujD^a?
+ +K^* + ^iVi+ +^,Vo)^+ —
be formed, in which the coefficient of any power of a; is
the same as in the product of the two first series; then
P will be a convergent series equal to UxV, provided
(1) that the series u and V have all their terms positive,
or (2) that the series U and V would not lose their con
vergency if the signs were all made positive*.
* This Article, and in fact the whole of this Chapter, is taken with
slight modifioationB from Caachy*B Analyse Algibrique,
CONVBRGENCY AND DIVERGENCY. 329
First, suppose that all the terms in U and V are
positive.
Then U^x F^ = P^ + terms containing a?'' and
higher powers of x. Hence U^ x V^ > P^.
Also Pjh, = CT, X F^ + other terms. Hence P^>TJ^x F^.
Hence P^ is intermediate to J7, x F^ and U^x V^.
Now, the series Z7and F being convergent, U^ and U^
both approach indefinitely near to 17, also F,„ and F^ both
approach indefinitely near to F, when n is indefinitely
increased. Hence U^ x F^ and CT^ x F«, and therefore
also P^ which is intermediate to them, will in the limit be
equal to CT x F. Hence, when all the terms are positive,
P=UxV.
Next, let the signs in the two series be not all positive,
and let ZT and F be the series obtained by making all the
signs positive in U and F; and let P' be the series formed
from U' and F in the same way as P is formed from
t/^andF.
Then U^ x F,„ — P^ cannot be numerically greater
than U'^ X V'^^P*^^ for the terms in the latter expres
sion are the same as those in the former but with all the
signs positive.
Now, provided the series U and F do not lose their
convergency when the signs of all the terms are made
positive, it follows from the first case that TT^ x V'^ — P'^,
and therefore also U^^ x V^ — P^ which is not numerically
greater, must diminish indefinitely when n is increased
without limit. Hence the limit of P^ is equal to the
limit of ?7j„ X F^ ; so that P must be a convergent series
equal to the product of U and F.
If the series U and F are convergent, but are such
that they would lose their convergency by making the
signs of all the terms positive, the series P may or may
not be convergent; and, when P is not convergent, the
relation UxV=P does not hold good, for P has no
definite value and cannot therefore be equal to tTx F,
330 CONVERGBNCY AND DIVERGENCY.
although the coefficient of any particular power of x in
the series P is always equal to the coefficient of the same
power of ^ in the product of the series U and V*.
281. If the two series
% + a^x '\ ajx? {• ajxf + ,
and h^'\\x\'h^ '\hjx?+ ,
be equal to one another for all values of x for which they
are convergent ; then will a^ = 6^, a^ = \, a, = 6,, &a
For if the series are both convergent, their difference
will be convergent Hence
for all values of x for which the series is convergent
The last series is clearly convergent when a? = ; and
putting a? = we have a^ — ft^ = 0. Hence a^ = 6© •
* We now have
^K^+K^J^+(«8^)^' + }=o (ii)
Now for any value a?, for which the series in (i) is
convergent, a, — 6j + (a, — 6j) ajj + is equal to a finite
limit, L^ suppose.
Hence (ii) may be written x^ [a. — 6^ + a?j i J = ; and,
since this is true for all values ot x., however small, it
follows that a,— frj, must be numerically indefinitely small
compared wito L^ ; that is, a^ — \ must be zero. It can
now be proved in a similar manner that a, — 6, = 0,
O3 — ft, = 0, &c.
Hence if two series which contain x be equal to one
another for all values of x for which the series are conver
gent, we may equate the coejfflcients of the same powers of x
in the two series.
The particular case of two series which have a finite
number of terms was proved in Art. 91.
* It can be proved that P is convergent if either U ox F is absolutely
convergent. See Chrystal's Algebra^ Part 11., p. 127.
EXAMPLES. 331
EXAMPLES XXVI.
Determine whether the following series are conveigent or
divergent:
1 1 1 1
^* 1 . 2 "*■ 3 . 4 "^ 5 . 6 "^ ••• "*■ (2n4. 1) (2n+ 2) "*■ 
1 1 1
3 3^ 3.4.5 3. 4...(w + 2)
^ 4 "^ 4 . 6 "^4 . 6 . 8 "*■•••'*■ 4 . 6...(2n + 2) "*" •"
. 3 3^ 3.5.7 3. 6.7.,.{2n + l)
^ 4'^4.7'^4.7.10"^'*"4.7. 10...(3u+ l)"*" *••
 I K3 1.3.5 1.3. 5. ..(2711)
• 3'^3.6'^3. e.O"^**'"^ 3. 6. 9.. .3m '^ '"
oil 1 1
X xh 1 x+ 2 x + 3
7 ^ 1 1
I +x 1 + 2a; 1 +3a;
1 1 1
• 1 +a;'^l+aj«"*' l+a;» "*"•••
Q 1 oj as" of
l+fc 1+ar l+oj* 1+05*^
1 1 1 1
1 .1/ .Er flj*
^^ rr2 "^ 273 "^ 371 '*■••*'*■ (n+ 1) (n + 2)^ •'*
12. l.f^+  — S •••♦'(l)"i + •••
1+a l + 2a ^ ^1+na
^32 EXAMPLEa
15. + jr— + «—+...
« 1 w m'
16. =^ + + 5+...
aj+1 a5 + m 05 + m"
(l + a)(l+6) (2+a)(2+6) . (« + a)(n + S) .
1.2.3 2.3.4 "^«(n+l)(n+2) "*■•••
1ft 1 2 3 n
'■''• 1+72 + 1 + 2^3 ""l +3^4'' •••■'i + mV^Ti''
"• 2 + ^2 "^3+ 3*4 + ^4 « + V«
21. (^21)4.(^52)+... +(^/;^N^n) + ...
1* "*" 3» "^ S*"*" '•• ■*" (271^1)*"^ •••
^, 2 4iB 6aj» 2naf
2 5 10 w*+l
oA 3 1 1 , 2?»5 ,_,
4 a Iz 7i8_5/j
25. Shew that the series
111 _\
ra;"^2».a;'^3»aj'*''""^» "*"•••
is convergent for all values of 05, except only when 05 is the
square of an integer.
rTo^,
26. sf77^4)
\^(nl) ^n)
27. S^/*{N/(^l)2^(7i2) + ^(n3)}a:.
CHAPTER XXIL
The Binomial Theorem. Any Index.
282. It was proved in Chapter xx. that, when n is
any positive integer,
(l + a;)=l+rw; + ^?^^^aj»+
I n(nl)...(nr + l) ^r ,
We now proceed to prove that the above formula is
true for all values of n, provided that the series on the
right is convergent
When n is a positive integer the above series stops, as
we have already seen, at the (n + l)th term; but when
71 is not a positive integer the series is endless, for no one
of the factors ri, n — 1, w — 2, &c. can in this case be zero.
It should be noticed that the general term of the
,. ., . , n(7i— l)(n — 2)...(7t — r + 1) 
binomial series, namely — ^^ ^^ r^ — ^ — ay,
n
cannot be written in the shortened form . — r= — of unless
\r \n — r
n is a positive integer; we may however employ the
notation of Art. 241, and write the series in the form
334 BINOMIAL THEOREM. ANY INDEX.
r
283. Proof of the Binomial Theorem. Represent,
for shortness^ any series of the form 1 + 77^^ + 70^^;*+ •••
[1 2
+nr^+ ... by/(m). Thus
/w=i+jir^+jF^+ +7^+ >
/(~)i+j?+f^+ +g
and
Now the coefficient of of in the product /(m) xf(n) is
I T^ 111 o lo "•" • •• "■" i i • • • . ' I f
that is
If \r ]
And, by Vandermonde's Theorem [Art. 249 or 261],
this coefficient is equal to ^^ , which is the coefficient
\r_
o{ of in f(m + n).
Thus the coefficient of any power of w in f(m + n) is
equal to the coefficient of the same power of x in the
product /(?m) x/(n); also the series /(m), f(n) and
f(m + n) are convergent, for all values of m and n, when
ic is numerically less than unity [Art 278]. It therefore
follows from Art. 280 that
f(m)xf(n)^f(m + n) (a),
for all values of m and n, provided that w is numerically
less than unity.
BINOMIAL THEOREM. ANY INDEX. 335
Now it is obvious that /(O) = 1, and that /(I) = (1 + a?) ;
we also know that if r be a positive integer fir) = (1 + xj.
Hence, by continued application of (a), we have
f{m) x/(n) y^fip) X ... =/(m + n) xf(p) x ...
=/(m + rn^ + ...).
Now let m = n =p =... = , where r and 8 are positive
integers; then taking s factors, we have
But, since r is a positive integer, /(r) is (1 +a?y;
.. (n^)'=/(3.
This proves the Binomial Theorem for a positive
fractional exponent: the theorem is therefore true for
any positive index.
And, assuming that the binomial theorem is true for
any positive index, it can be proved to be true also for any
negative index. For, from (a),
/( n) x/(n) =/( n + ») =/(0).
Hence, as/(0) = 1, we have
•^^"" ^^ "Too " (iT^ ' ^^^^^ ^ ^^ positive,
= (1 +«?)*.
Hence (1 + a?)"* =/(— n), which proves the theorem
for any negative index.
284. Euler'8 Proof. Euler's proof of the Binomial
Theorem is as follows.
336 euleb's proof.
Represent, for shortness, any series of the form
1.2 It
hj/{m): thus
f(m) = l+mx + ^a? + ... + 1^af+ (i),
/(n)=l+fuv + ja^ + ... + "' of + (ii),
and,
/(m + w) = 1 4 (m+n)x + ^ — 12 "^ "• + \ ^+ •••
Now, if the series on the right of (i) and (ii) be multi
plied, and the product be arranged according to ascending
powers of x, the result must involve m and n in the same
way whatever their values may be. But, when m and n
are positive integers, we know that/(m) is (1 +aj)"', and
that/(w) is (1 +a?)*, and the product /(m) x/(n) is there
fore (1 + a?)*""**, which again, as m + w is a positive integer,
isf{m + n). Hence when m and n are positive integers
the producty(m) xy(w) is/(m4ri); and, as the form of
the product is the same for all values of m and n it follows
that
/(m) xf{n) =f{m + n) (a),
for all values of m and n, [See however Art. 280.]
From this point the proof is the same as in Art. 288.
Ex. 1. Expand (1 + «)!.
Put n=  1 in the above formula; then we have
■■.+<^)<f <^)^4
=l»+a»a!»+ +(Vfi^+
BINOMIAL THEOREM. ANY INDEX. 337
This example illustrates the necessity of some limitation in the
value of X} for we know [Art. 229] that loj+o;* is not
equal to = unless x is between  1 and + 1.
Ex.2. Expand (!«)*.
We have
(i»)«=i^(2)(.)^(4L:a(,).^ (^HBH4) ^_^^.
+ ^(2)(8) .■'(Ma)(.^),^
II.
= l+2ar+3a;«+4a:»+ + {r+l)x^+
Here again it is dear that the result cannot be true for all value$
of X ; if «=2, for example, we should have
1=1 + 2.2+3.22+4.23+ ,
which is absurd.
Ex. 3. Expand {l+x)i.
We have i^ + ^)^==^+l^ + ^^^'''^ ^^ i%\ ^^ ^'+
the general term being
Hence (l+*)*=l + 5^0 '' + 27176*'+
^^ ' 2.4.6...2r * ^
Ex. 4. Expand (1  x)'K
Wehave (l»)*=l + (i)(a;)+ \ V^ M ^)''+
iJKJbL^,....
Hence
1 (^)'
/I X* 1.1 .13 a. 1.8.6 (2rl) „.
(l«) i=l+ix + ^a;« + + 2.4.6...:..2r ^ ^
All the terms are positive, for in the general term there are 2r
negative factors.
S. A. 22
338 BINOMIAL THEOREM. ANY INDEX.
Ex. 5. Expand (a>  Za^x)^ acooifding to ascending powers of x,
(«.  8a»»)*= a. (l  ^)\ *=«. (l  1)*
6 2 ^ l( A\
[•*i(?)*H(?y*'4#(?)'
♦ J^');(i"') (.i)v ]
 r, 5 X 6.2 /a?V 6.2.1 fxY
6.2.1.4.7...(3r8) /f V . 1
After the second, all the signs are positive ; for in the general term
there are r  2+r, that is an even number, of negative factors.
285. The (r + l)th term of the expansion of (1 +a?)*
is obtained from the rth by multiplying by a?,
r
that is by ( — 1 H J x. Now — 1 i is always
negative if n 4 1 is negative ; and, whatever n + 1 may be,
71 + 1 .
— 1 H will be negative for all terms after the first for
which r>n + l.
Hence, if w be positive, the ratio of the r + 1th and rth
terms will be always negative when r>n + l. The terms
of the expansion of (1 + a?)* will therefore be alternately
positive and negative after r terms, where r is the first
positive integer greater than w + 1.
If X be negative, the ratio of the (r + l)th and rth
terms will be always positive when r > n + 1. The terms
of the expansion of (1 — a?)" will therefore be all of the
same sign as the rth term, where r is the first positive
integer greater than n + 1 ; and, as a particular case, all
the terms of the expansion of (1 — a?)" are positive when n
is negative.
GREATEST TERM. 339
For example, all the terms in the expansion of (lx)' are of the
same sign as the rth, where r is the integer next greater than  + 1,
00 that r is 8. Also, after the ninth, the terms of the expansion of
(l+x)^ are alternately positiye and negative.
286. Greatest Term. In the expaDsion of (1 ± xy
by the binomial theorem, we know that the ratio of the
(r + l)th term to the rth is ± x, that is
T a:fl ]; we also know that x must be numeri
cally less than 1, unless n is a positive integer.
First suppose that w + 1 is negative, and equal to
— m. Then the absolute value of the ratio of the (r + l)th
term to the rth term is a; ( 1 H — J. Hence the rth term
is = (r + l)th term according as x (l \ — ) = 1 5 that is,
,. > mx ^, ^ . > — (l'hn)x
according as r = z , that is = — \i .
° <1— a? < 1 — x
Hence, if — ~ — be an integer, r suppose, the rth
term will be equal to the (r + l)th term, and these will
be greater than any other terms. But, if — ^=— — —
be not an integer, the rth term will be the greatest when
r is the integer next above — ^ —.
^ lx
Next, suppose that w + 1 is positive, and let k be the
integer next greater than w + 1, Then, if r be equal or
greater than k, 1 will be negative and less than
unity; hence, as x must be less than unity, each term
after the kth will be less than the one before it, and
therefore the greatest term must precede the kth. And
since, for values of r less than n + 1, 1 will be
r
32— 2
340
posttiTe; the rih term will be =(r + l)tli acocmliiig as
Hence, if ^ ., be an integei; r sai^Kwe^ the rUi
term will be equal to the (r+ l)tli, and these will be
greater than any oAer tenna Bat^ if ^r= — ^ be not an
int^er, the rth t^rm will be the greatest when r is the
mt^;^ next above ^^ — —j •
Ex. 1. Find tibe grastert torn in tibe eipanrimn <rf (l«)^, wbm
ir=. HeiB «+lk negative, and ^^^ii^=ii==4. Heneeihe
fourtii end fifth temu aie equal to one another, and aie greater
than anj other terms.
Ex. 2, Vtad when tibe expansion of (lx)~V begins to eonyerge, if
8
'=!•
Here«+li«iie(pMitB,aiid— ^p±^=^^=2H. Heneeihe
coDvetgenee begine after the 23rd tenn.
Ex« S, Find the greatest tenn in the expamdon of (a+x)^, when
4«s8a.
Since (a+z)'^=a^ ll + j , the greatest term required is the
/ «\V
term corresponding to the greatest term in (1 + ) . Now
(n+l)ff 1+ j=j •7^2~5» hence r must be the integer next
greater than ^ , so that the 5th term is the greatest.
EXAMPLES XXVII.
L Find the general term in the expansion of each of the
following expressions by the binomial theorem.
(i) (1*), (ii) (1*), (iii) (!«).
EXAMPLES. 341
(iv) (l+x)^, (v) (1+a:)*, (vi) (l+o:)*
(vii) (l.5a;)t, (viii) (l5a;)», (ix) (laj)"?,
(x) (2a+3iB)*, (xi) (a«  2aa?)*, and
(xii) (47aj)f
2. Find the first negative term in the expansion (i) of
(1 + laj)"^, and (ii) of (1 + %x)^.
3. Find the greatest term in the expansion of (1 +05)""
when 05 = J.
4. Find the greatest term in the expansion of (1 fa;)""
when 05 = f.
5. After what term will the expansion of (1 — 05)H* begin
to converge, when 05 = ^?
6. Shew that the coefficients of the first 19 terms in the
19 — 21a;
expansion of t:j r, are all positive, and that the greatest
of them is 100.
7. If ttj, a,, a,, a^ be any four coefficients of consecutive
terms of an expanded binomiiJ, prove that
a a 2a
8. Find the general term in the expansion by the binomial
theorem of each of the following expressions according to
ascending powers of x:
(iv) (a + x)i {a  a;)~4, (v) (a + x)* (a  05)"*, and
(vi) (a^xy{a^x)'K
9. Shew that the coefficient of a^ in the expansion of
(1 +ar»)»(l  «;»)• is 2w.
10. Shew that the coefficient of 05^ in the expansion of
(1 + 2a;)»(l x)" is 27 (n 1), w^: 3.
342 SUM OF COEFFICIENTS.
287. Sum of coefiBLcients. The sum of the first
r + 1 coefficients of the expansion of (1 — x)* can be ob
tained as follows.
We have
(i^r=ipia;+^^«...+(iy%^+...,
also (1 —«?)"* = ! +x + a^+ ...+af'^ ...
From [Art. 281] the coefficient of of in the product
of the two series is equal to the coefficient of of in
(1 — xy X (1 — a?)"*, that is in (1 — ^)*"*; hence we have
^ IJ. ^[2 ^^ ^^ \r
= coefficient of af in (1  xy' = ( ly ^V;^ .
Similarly, ii <f>(x) = a^ + a^x¥ a^ + ... + aX + "$ ^^^
sum a^ + a.+... + ar will be the coefficient of of in ^^ .
Thus, to find the sum of the first r + 1 coefficients in the
expansion of (f> (x), we have only to find the coefficient of
of in the expansion of y^ .
Ex. 1. Find the siim of the first r ooefficients in the expimmon of
(la?)». Atu. Jr(r+l)(r+2).
The sum required is the coefficient of af^^ in (1 a?)*.
Ex. 2. Find the sum of n terms of the series
1.2.8 + 2.8.4+8.4.5 +
Sinoe(l«)*==io[l*2.8+2.8.4a; + 8.4.5«»+ ];the
sum required =6 x sum of the first n ooefficients in the expansion of
(1 »)*= 6 X coefficient of a;»iin (1 »)•= jn(n+l)(n +2) (n+3).
Ex. 8. Find the sum of the first n+r ooefficients in the expansion of
BINOMIAL SERIES. 343
The sum required = coefficient of a?*+*'"^ in the expansion of
ji^ll^. Now(l+a;)*=(2r^)*=2*n.2*i(la;)
^n^tj^ 2*a (1  «)»+ higher powers of (1  x),
Hcnco <^ + ^)*  ^ J^?!± + M!Lia?!l%an integral
expression of the (n  3)th degree.
The coefficients of x*+^i in (!«)», (1a?)"^ and (lx)i re
spectively are 5 (n+r) (n+r+ 1), n+r, and 1 ; hence the coefficient of
_^^. . (1+0?)*.
a*"^^^ in 7 ^i 18
2»i(n+r)(n+r+l)2*in(n+r) + 2*8n(nl).
Ex. 4. Find the sum of n terms of the series
Atu. (2nl)I/nl(nl)I.
288. Binomial Seriei. Series which are derived
from the expansion of (1 + w)* by giving particular values
to X and n are of frequent occurrence: it is therefore of
importance to be able to determine at once when a given
series is a binomial series.
The case in which the index is a positive integer needs
no remark.
When the index is a negative integer, we have
. w(reHl)...(w + rl) _, ,
and it should be carefully noticed that this expansion can
be written in the form
(!«?)"**= — 3Y[1.2...(7il) + {2.3...7i}a? + ...
...+ K^ + l)...(r + nl)}a?''+...].
344 BINOMIAL SERIES.
When the index is fractional, — p/j suppose, we have
fP(£+S^+M (ij + (A).
Here we notice that (i) there is an additional factor
both in the numerator and in the denominator for every
successive term, (ii) the successive factors of the numerator
are in an A. P. whose common difference is the denominator
of the index, (iii) the successive factors of the denominator
are 1, 2, 3, 4, &c., or multiples of these.
Bearing in mind the above laws, there will be no
difficulty in determining the expression which will pro
duce a given binomial series.
Ex. 1. Find the Bom of the series
1 . 1.8 . 1.8.6 . . ....
S + O + sTO* *«"^<y
Writing the series in the fonn
1 1 1.3 1 1.3.5 1 ^
1^*8 2'8>'** 8 'a^"*" *
we see from (A) that it is ohtained from the expansion of (1  x)~^
by giving to x the value found from » = q .
1 8
(2\~i 12 2*2 /2\*
Ex. 2. Find the sum of the series
 2 . 2.6 , 2.6.8 . X . « .x„
^•^6 + 6712 + 02718+ tomfimty.
Writing the series in the form
+ [1*6+ 2 62+ [3 •6» +
we see from (A) that it is obtained from the expansion of (1  x)
«\~i
BINOMIAL SERIES. 345
X 1
by giving to x the valae 0=2 • Hence the sum required is
(.!)•*.
Ex. 3. Find the sum of the series l+j^ + ^l^^+ to
infinity.
In this ease the factors of the denominator, although multiples of
1, 2, 8, 4, &c., do not begin at the beginning. Additional factors
most tiierefore be introduced in the denominator, and corresponding
additional factors in the numerator. We then have
(,5)(l)3 1 (g)(l)8.7 1
8 e*"*" 4 6*"*"
Now the terms of this latter series are terms of (A), if 9^4, jis  5,
*^^4 = 6
We can therefore find the required sum, as follows :
/ 4\t 6 1 . 6.1 1 . 5.1.8 1 I 5.1.8.7 1
5 r8 8.7 8.7.11 .
'*'6.12Ll8"^18.24'*"l8.24.30'*' J'
•*• \^) "^"6'*'72'*"72^'
4\t . 6 1 6.11 5.1.8 1 5.1.8.7 1 .
6 _5_ 6 rS . 3.7 . 8.7.11
" 6"*'6.12
l\i .5 6.6
Whence 5 = i {8 4^27  17}.
Q ft 5 3 6 7
Ex. 4. Find the sum of the series 7 + ^^ + . ' * + to infinity.
4 4.0 4.0. 12
[From^liy 1. Ant. 2^/21.
Ex. 6. Find the sum to infinity of the series
1 1.8 1.8.6
2»8~2*(4'*' 2»[6 "
23 2
[From (1 + 1)*]. ^»« 24 " 8 '^^*
■^ n ox, XI. X , 1 14 1.4.7 1.4.7.10 . . .
Ex.6. ShewthatlH^ + ^.f^g^4 ^ 3^^^^^^ + to m
^ ..  2 2.5 2.6.8 2.6.8.11 . * ,. x^
^*y = l + 6*6n2 + 6n08^ 6712718724 + ^^^^^
[s,«(>f)'.(.i)^.
846 THEOBEMS OBTAINED BY EQUATING COEFFICIENTS.
289. We know from Art. 281 that if any expression
containing x be expanded in two diflferent convergent
series arranged according to ascending powers of x, the
coeflficients of like powers of x in the two series will be
equal. By means of this very important principle many
theorems can be proved.
Ex. 1. Sliew that, if » be any positive integer,
n« n«(n«lg) n»(n«l«) (»«2»)
«T !> /I x« ^ .*»(nl)  n(nl) (n2) , .
We have (la;)*=liia?+— ^j— 5— ' a?« ^ , '^^ — a^^
■K_l),»("l)("i^)^.
1 . 2.;.ll
Also, provided a; > 1, we have
/ rv^. , „1 , n(n+l) 1 n(n + l)(n+2) 1
V'x) =^+'*i+"r:2"^+ 1.2.3_ i5+
n(n+l)...(n+nl) 1
"*" l,2...n a;*"*"'""
Hence ln«+^^ +(!)« ^^^, '
is equal to the coefficient of 0:^^ in (1 x)*^ x ( 1  J , that is equal
to the coefficient of «* in (  l)*ic*, which is zero. [See also Art.
261, Ex. 3.]
Ex. 2. Find the sum of
(n+l)+n. + (nl)2^ + + 1. 2.4.6...2n *
[Equate coefficients of x* in (1  a;)"' x (1  x)^ and in (1  a?)"*.]
6.7...(2n+8)
^^ 2.4...2n ■
Ex.3. Shew that l3n+^i5:ii^^^^ = (1)*.
We have 3p^, == J— ^^ = j^^^^^j .
Hence (l+a:){la?+«^... + (l)"a5»*+...}
= l+a;(lx)+a:?(la;)2+...+x»Hi(ia.)8iHi + .,.
The coefficient of {b»*+i on the left is (  1)\
EXPANSION OF MULTINOMIALS. 34}7
The terms on the right which give a^+^ are
aj?*+i (1  a;)'^+^ + a:^ (1  a?P + a5®*Ml  a;)**! + . . . ;
and henoe the coefficient of a?'^^ will be found to be
1 o^ A^m^^) (3n2)(3n3)(3n4) .
l3»+ j^ j23 — .+
290. Ezpaniion of Multinomials. Any multi
nomial expression can be expanded by means of the
binomial theorem.
Since (j) + qx + rx^+ ., .)** may be written in the form
p*(l +«?+ a^ + ... j , it is only necessary to consider
expressions in which the first term is unity.
Now in the expansion of {1 + ax + baf^ ^ ca? + .,.Y,
that is of {1 + (ax h bar^ + caf + .,.)}'', by the binomial
theorem, the general term is
n(nl)(n2)...(nr + l ) j^^^^^
also in the expansion of (ax + ba? + ca^ + ..,y, r being a
positive integer, the general term is by Art. 262
Ir
T—rw=i a'^bPcy . . . aj»+2^+»y+...,
a^7...
where each of a, y8, 7,... is zero or a positive integer, and
oi + l3 + y+ ...=r.
Hence the general term of the expansion of the
multinomial is
r.(nl)(n2)..(nr + l) ^.j^,....^^2,^a,...,
a [^ _7...
To find the coefficient of any particular power of x,
say of a^, we must therefore find all the different sets of
positive integral values (including zero) of a, /8, 7,...
which satisfy the equation a f 2/8 + 87 4 ... = A?; the cor
responding value of r is then given byr=a + /8 + 74...,
and the corresponding coefficient is found by substituting
348 EXPANSION OF MULTINOMIALa
in the formula for the general term. The required coeffi
cient wiU then be the sum of the coefficients corresponding
to each set of values of a, fi, 7....
Ex. 1. Fmdtheooeffioieiitof ffBin(la;+22'3a;')~'i*
The values of a, ^, 7 which satisfy a+2^+37=5 will be found
to be 0, 1, 1; 2, 0, 1; 1, 2, 0; 3, 1, 0; and 5, 0, 0. The cor
responding values of r will be 2, 3, 8, 4 and 5 respectively; and the
corresponding coefficients will be
^, (i)(i)(mD(l) , ,„
.... 9 45 15 35  63
**^**"2' W 4' 16 "'^ 256
31
Hence the required coefficient is rr= •
256
291. From the above example it will be seen that
the process of finding even the first six terms in the
expansion of a multinomial is very laborious; in many
cases, however, the work can be much shortened, as in
the following examples.
Ex. 2. Find the coefficient of x^^ in the expansion of
(l+a?+a?«+«»+«*)9.
We have (l+a?+«»+«»+a^)9= ^lr^\"*={ix)i(l _«»)«
= (l2aj+«*)(l + 2a!5+3a?W4.4a.w+...).
Hence the coefficient required is zero.
Ex. 3. Find the coefficient of o^ in the expansion of (1 + x + jb* + a^Y'h
Wehave(l+x+«»+a;3)i= ^— s= l^,
=(la;)(l+a?*+«8+...+aj*"+...).
COMBINATIONS WITH REPETITIONS. 349
Henoe the coefficient of a^ is 1, the coefficient of s^^ is  1,
the coefficient of j^^^ is zero, and the coefficient of v^^ is zero.
Thus the coefficient of a^ is 1 when n is of the form 4r, it is ~ 1
when n is of the form 4r + 1, and it is zero when n is of eith» of the
forms 4r +2 or 4r+3.
Ex. 4. Find the coefficient of oc** in the expansion of
(l + 2a;+8aB^+4a^+ to infinity)*.
Since l + 2^+8a^+ = (l~a;)~^ the required expansion is
that of (1  «)""*•; lie coefficient of x^ is therefore •
2n(2n+l)...(2n>rl)
ii
292. Combinationi with repetitioni. The number
of combinations of n things a together of which p are of
one kind, ^ of a second, r of a third, and so on, can be
found in the following manner.
Let the different things be represented by the letters
a, 6, c,...; and consider the continued product
(l+aa?+aV+...+a''aj')(l+6a:+..,+6V)(l+ca?+...c*'aO.
It is clear that all the terms in the continued product
are of the same degree in the letters a, 6, c,... as in ^; it
is also clear that the coefficient of of" is the sum of all the
different ways of taking a of the letters a, 6, c,... with the
restriction that there are to be not more than jp as, not
more than g &'s, &c. ; so that the coefficient of of in the
continued product gives the actual combinations required.
Hence the nwmber of the combinations will be given by
putting a = 6 = c=... = l. Thus the number of the com
binations of the n things a together is the coefficient of
a^in
(l+a? + a;* + ...+a;»')(l+a?+...+aJ')(l + a?+... +ar)...
PermutationB. The number of permutations of the
n things a together being represented by P., it is easily
seen that
jl+^+2 + ... + i^}l+il + j2+ +i^}x
P P P
350 PERMUTATIONS.
For la X the coefficients of ic* in
ax a^a^ a^a^)
"5! """IF "*"***"*" Iff
{
{ bx
1+ 2 "^•••■^ ?j "■
is the sum of all possible terms of the form
la
a'6*...
i
m ...
for which Z + m + ... = a, and the number of permutations
a together formed by taking I of the a's, m of the 6's, &c. is
I \fn ... *
Ex. 1. Find the number of combinations 7 together of 5 a*s, 4 5's
and 2 c's.
The nmnber required is the coefficient of x^ in {l+x^.^+sfi)
(1+X+ ...a^)(l+a;+x»), that is in (1  x^J (1  aH^) (1  a?*) (1  «)».
Bejecting terms of higher than the seventn degree in the continued
product of the first three factors, we have
(la;»«»aj«)(l + 8aJ+6a:»+10«»+16ai*+21a5»+28aj« + 8ea^+...);
and the coefficient of x^ is 86  15  6  8= 12.
Ex. 2. Find the total number of .ways in which a selection can be
made from n things of which p are alike of one kind, q alike of a
second kind, and so on.
The total number of the combinations is the sum of the coef
ficients otx\ 05*,..., a* in (l+x + ...+a;'»)(l + a; + . ..+««)...; and this
sum is obtained by putting x=l in the product and subtracting 1
for the coefficient of aP, Hence the required number is
(l, + l)(g + l)...l.
The above result can, however, be obtained at once from the
consideration that there are p+1 ways of selecting from the a*8,
namely by taking 0, or 1, or 2,... or p of them; and, when this is
done, there are q + 1 ways of selecting from the 6*s ; and so on.
Hence the total number of ways, excluding the case in which no
letter at all is selected, is ( p + 1) (g + 1) . . .  1. [Whitworth's Choice
find Chance, Prop, xiii.]
HOMOGENEOUS PBODUCTS. 351
Ex. 8. A candidate is examined in three papesrs to each of which m
marks are assigned as a maximum. His total in the three papers is
2m; shew that there are ^(m+l)(m + 2) ways in which this may
oconr.
The number of ways is the coefficient of a^ in (1 + « + a:* + . . .rt")*,
that is in (1  aj"»+i)» (1  x)8= (1  3a^+i + ...) x
{1.2+2. 8a;+...+m(i»+l)x**i + ...+ (2m+l){2m+2)a^{ .,.].
Hence the number required
= i{(2ifi+l)(2m+2)8in(m+l)}=i(m+l)(m+2).
Ex. 4. Shew that the number of permutations four at a time which
can be made of n groups of things of which each consists of three
things like one another but unlike all the rest is n^  n.
The number required is equal to 4 x the coefficient of x* in
(^ X a^ x^\^
293. Homogeneoui Products. We have already
[Art. 250] fouDd the number of homogeneous products of r
dimensions which can be formed with n letters, where each
letter may be repeated any number of times. We now
give another method of obtaining the result. Suppose the
letters to be a, 6, c,... ; then if the continued product
{l+ax + aV + aV + ...) x (1 + 6a? + 6V + 6V + ...)
x(l +ca?HcVf c*i»^ +...).. .
be formed, the coeflficient of a?*' will clearly be of r dimen*
sions in the letters a, b, c,..., and will be the sum of all the
possible ways of taking r of the letters*. Hence the
number of the products each of r dimensions will be given
by putting a=6=c=,.. = l in the continued product.
Thus the number required is the coefficient of of in
{1 + x + a^ + ,,,y, that is in (1 — a?)"*. Hence
n (n + 1). . .(n + r — 1) [n f r — 1
This result can be expressed in the form ^H^ = n+ri^r
* An expression for the sum of the homogeneous products will be
found in Art. 300, Ex. 4.
352 EXAMPLES.
Cor. The number of terms in the expansion of
In + r— 1
(a, + a, + a,+ ... + a,y is ' ^^^.^ .
294. We shall conclude this chapter by solving the
following examples.
Ex. 1. Find J\^ hj the binomial theorem, to six plaees of decimals.
=4 {1  0626  001953  0001220  0000095  0000010}
=3741667.
Ex. 2. Shew that, when x is small,
(l3x)!+{l4a;H ,8 ...
'—I — i '^ = 1 + ;r « approximately.
(l3a;)4+(l4«)i ^
Since x is small, its square and higher powers may be rejected ;
and when aU powers of x except the first are neglected the given
expansion becomes equal to
o 4
= (i+5»)(i+«r=(i+«)(i«')=i+
X,
Ex. 3. Shew that the integral part of
(V3 + 1)»»+J is (V8 + l)*»+i  (>/3  l)2»+i.
Since a/SI is a proper fraction, (^31)"*+' must also be a
proper fraction. It therefore follows that if (^3 + !)**+»  (^3  !)*•+*
be an integer, it must be the integral part of (/^3+ 1)^'*'^*
Now (^3 + l)«»+i  (V3  1)*»+^
= {3«^/3 + (2n+l)3*+^^^t^3*V3 + + (2n+l)V3+l}
{3«V3(2n+l)8»+ ^?^p^^^3*V3 + (2n+l)VSl}
=2U2n + l)3^ + < ^^\)^^;f^^> 3«>^ + + 1},
all the irrational terms disappearing.
EXAMPLES. 353
Since the ooeffioients of all the different powers of 3 in the last
expression are integers, it follows that (/^3 + 1)^"*"^  (/^3  1)**"*"^ is
an integer, and is moreover an even integer.
By the following method it can he proved that
(/^3 + 1)*^^  (V3  l)2*+i is an integer divisible by 2»+i.
Bepresent (^3 + l)**+i  (^3  l)^+i by I^^ .
Then Ii=2; and it will be found that 13= 20, and also that
U/3 + l)a+(V3l)«=8.
Hence
Bl2«+i = {(>/8 + l)*'+^(V3l)2»+i}{(V3 + l)2 + (V3l)2}
= (V3 + l)«*+3_(^3_l)2«.+«^.4{(^3^.1)2»i_(^3_l)2«i}.
•*• •^2n+S = B^2n+l ""^2111 (^)'
It follows from the last relation that I^n+s '^^ ^® <^ integer if
I^n+i and I^ni cure integers. Now we know that I^ and I^ are
integers ; hence by induction I^^i is always an integer.
The relation (A) also shews that 131,4.3 ^^^ ^ divisible by 2"^
provided I^^^ is divisible by 2»+^ and I^ni ^7 2*. Now we know
that Jj is divisible by 2^ and J3 by 2^ ; hence J5 must be divisible by
2^; and it will then follow that J7 must be divisible by 2^ ; and so on,
so that lan+i is always divisible by 2*+^.
Ex. 4. To shew that, if n be any positive integer,
a*n(a+&)»+^^^^\a+26)« = (6)«w..
Put ^T— for X in the identity proved in Art. 259, Ex. 3 ; then,
after reduction, we have
l2L^* *__Co Cj
T • • •
{y + a){y+a+h) ...(y + a+nb) y + a y+a+b
y+a+rb
Now expand the expressions on the two sides in powers of  .
n6* In 6*
Left side = j ^~ ; = rrr^ + higher negative
powers of y.
M«htBida=^(i+?)'....(irJ(i+»±':yV
• • ••* f
hence the coefficient of r^? on the right is
(l)*[Coa*Ci(a+6)*+ + ( Ifc^ (a +r6)* +...].
Hence S (l)%(a+r5)* is zero if fc<n, and is equal to
(l)»6»nif fe=n.
S.A. 23
854 EXAMPLES.
EXAMPLES XXVni.
1, Find the sum to infinity of each of the following
series :
13 1^3^ 1.3.5 3'
(i) l+i25^"*" 2 2«"^ 13 2''*"*"
11 13 1 1.3.5 1
(") 1 22"*'2.42^ 2.4.62*"^"'
r"\ 1 ^1 ^•'^1 ^'7.10 1
(ui) 1 + i 4 ■*" "l^^* ■*■ [3 4' "*" •••
3^ 3.5.7 3.5 .7.9
^'""^ 3.6"*'3.6.9"*"3.6.9. 12'*""*
3 3.4 3.4.5 3.4.5.6
^^^ 2.4"^2.4.6'*'2.4.6.8'*'2.4.6.8.10
2 2.5 2.5.8
^""'^ ^"^6"^ 6712^6. 12. 18 ■*■•••
, ..  3 3.5 3.5.7
/ N ± 4»12 4.12.20
(vui) 18'*'18.27'*"18.27.36"^"'
^^^' ^■*"9"*"9TT8"^9.18.27"^'*'
, , 1 1.3 1.3.5
9.18 9.18.27 9.18.27.36 '"
( '\ 1 1.3 1. 3.5
^^^^ 2. 4. 6 ■*" 2. 4. 6. 8 ■*■ 2. 4. 6. 8. 10 ■*"•"•
( "\ 1. '^'^Q 7.28.49
^^'^^ 72'*"72.96"*"72.96.120'*"*"
2. Shew that
a + b 1.2 \a + 6/ 5_
6 w(n+l)/ 6 V "^•'
1+w +— L_/( ^) + ...
a + 6 1.2 \a + 6/
1.3.5
•^2
EXAMPLES. 365
3. Shew that
(l+.r=2"fi«l:if + !^(lz5y
^ ' \ l + x 1.2\l+a;/
1.2.3 \\^x) ■*■••••/•
4. Shew that, if x be greater than  J,
35 _ a; 1 / a; Y 1_^ ( ^ \
7(^+1) " T+a ■*" 2 VTT^y "^274^11^/
.3.5 / X Y
.4.6 Vl+a:/
5. Shew that
(1  7?r = (1 + xy  27za; (1 + a:)'^ + 2yi(2n~2) ^^^ ^^y^^_ , , ,
6. Shew that
a^x n{n+\) /a — asY _ /^ "*" ^V
^oTx 1.2 \a + ft:/ ~ \ 2a: /
7. Shew that
(1 +a;)«" = (l +aj) + 7ia?(l +a;)» ^ 7i(n.^ 1) ^ ^^ ^^^,.,
8. Shew that, if a < 6,
2L2 / rx4K 4 a" ^5 a* 4.5.6 a* )
a6=(a + 6) _^ +_^j2^. + J.
9. Shew that
n + x (n+2a;)(wl) (w + 3a;) (n 1) (w2) _^
^"TT^"*" [2(l+aj)« [3(l+aj)« ■^•••" *
10. Shew that, if the numerical value of y be less than one
third of that of Xy
aj + y 1.2 \x + y/ 1.2.3 \x + y/
a;y 1.2 \xy/
23—2
■^ . . •
356
11. i^nd the valae of
■.(rl)n.(r2r^>(.3) "<7^]<^> ....
n—
tor terms.
12. Shew that^ if n be a podtiye integer^
n'(nl) n'(n'l')(n2)
(12 ^ 23
[3 [4
13. Shew that, if n be a positive integer,
" 1 2 ^ gjS
^ ' k r + 1 ^
14. Shew that if n be a positive integer <^ 4
1 ,^ .4.5 n(7il) 4.5.6n(nl)(n2) .
^"*^"^ 172 "172 r^73 r:T"3 + • "•
15. Shew that
l.w(w + l) + 2(wl)7t+3(n2)(7il)+...+n.l.2
= Yo w (w+ 1) (^ + 2)(w + 3).
16. Prove that
l.n(»+l)+^.(»l)n + ^^4^(«2)(»l)
w(n+l)(w+2) 2w+l
■" 17273 («3)(w2) + ... = 2— ;^ , _^^.
17. Shew that, if p, = i4^^:i!!lzi),
EXAMPLES. 357
,^ ^^ 1.3.5...(2r~l) , 5.7...(2r + 3)
^^' ^^^r= 2.4.6...2r >^^^= 2.4...2r P^^^^
that p^ ^Pr.^i +Pr»9s + ... + fi'r = i (^ + 1) (»• + 2).
19. Shew that
= I {2"^' + (!)•}.
20. Shew that ^^^ = (a + ft)""'  («  2) 06 (a + 6)«
21. From the expansion of (1 + 2a; + (c*)" prove that
2 n(2nl) ... 2«(2nl) (2«2) (2>t 3)
till El!
In \n \2n \2n
22. Shew that
n(n+ l),..(n + m 1) n(7i+ l)...(n + m — 4)
Im Im — 3
w(w 1) w(n+ l)...(w + w — 7) _^
1.2 m6
if m > 2n, and = 1 if w = 2n.
23. Find the coefficient of a?" in
24. Shew that, if a; be a proper fraction,
(la:)(la^)(la;*)(la;0..." ^^ +«^) (^ + ^') (^ +«^)
25. In how many ways can 12 pennies be distributed
among 6 children so that each may receive one at least, and
none more than three )
358 EXAMPLES.
26. There are n things of which p are alike and the rest
unlike ; prove that the total number of combinations that can
be formed of them is ( jt) + 1) 2""'  1.
27. Shew that the number of ways in which n like things
can be allotted to r different persons, blank lots being admis
sible, is .+,.^(7,.,.
28. Shew that the number of combinations n together of
2n things, n of which are alike and the rest are all different,
is 2".
29. The number of combinations n together of 3n things,
of which n are alike and the rest all different, is
2'""^ + 2nl/n [rtl.
30. A man goes in for an examination in which there are
four papers with a maximum of m marks for each paper ; shew
that the number of ways of getting half marks on the whole is
^(m+l)(2m'+ 4/11 + 3).
31. Find the coefficient of a* in (1  2a; 2a^)K
32. Find the coefficients of x^ in the expansions of
(l+x+a^ + a^ + x^y and (1 + a; + a:* + a* + jb* + a;*)'.
33. In a shooting competition a man can score 5, 4, 3, 2, 1
or points for each shot. Find the number of different ways
in which he can score 30 in 7 shots.
34. In how many ways can 20 be thrown with 4 dice, each
of which has six faces marked 1, 2, 3, 4, 5, 6 respectively )
35. Find the coefficient of x"" in the expansion, according to
ascending powers of as, of (4a' + Goa; + 9a:^)~\
36. Shew that the coefficient of a:^ in the expansion of
IS 2flli 1.
(1 + aj + a^)
37. Shew that the coefficient of a;' in the expansion of
(1 + 2aj + 3a;'+ ...)» is J (r + 1) (r + 2) (r + 3).
EXAMPLES. 359
38. Find the coefficient of a;** in the expansion of
{1 . 2 + 2 . 3a; + 3 . 4aj« + ... to infinity}".
39. Find the coefficient of a?' in the expansion of
(1.2 + 2. 3. 2a; + 3. 4. 2V + + (^ + i) (n + 2) 2"a;"
+ ... to infinity)".
40. Shew that the coefficient of af in the expansion of
(1 + a; + 2a:» + 3a;» + ...)• is Jr (r^ + 11).
41. Shew that if pq be small compared with p or q,
then will
yp ^ (n+\)p + {n^\)q ^
y q (wl)jt) + (7i+ l)g ^'
42. If (6 V6 + 14)*"+* = ir, and i^ be its fractional part;
then wiU NF=^ 20"+^
43. If (3 ^3 + 5)"''^* = /+ i^, where / is an integer and F
a proper fraction, then will ^(/+ ^) = 2""+^
44. Shew that the integer next greater than (3 + ,^7)""
is divisible by 2"**^
45. If m be a positive integer, the integer next greater
than (3 + J^y is divisible by 2\
46. Shew that the general term in the expansion of
1 +x\y + xy
1 +x + y
\m + n^2
ia ( 1)+ ' xy
m— 1 n—l
47. Shew that the coefficient of a;' in the expansion of
X . f, r'l" (r»  1«) (r«  2») ,
J. ^ IS r U + — r^— c + ^^ 4^^ c'
• • • I •
(r'l')(r'2°) (r'3') ^. ^
II
860 EXAMPLES.
48. Shew that
1.2.n + 3.4^^> + 5.6!^<!^^^
+ (2n3)(2n2).t» + (2nl)2i».l=2n".
49. Shew that the coefficient of of^*^' in the expansion of
50. Shew that the coefficient of a^"^^* in the expansion of
51. Shew that
?*"  w (w  2)" + ^V~ ' (n  4)"  ... to n + 1 terms
= 2.4.6.8...2n.
52. Shew that
a"*'  w(a + 6)*' +^5^?^^ (a + 26)^* ...
= J (n+1 (2a + n6) ( 6).
53. If three consecutive coefficients in the expansion of
any power of a binomial be in arithmetical progression, prove
that the index, when rational, must be of the form ^ — 2,
where ^ is an integer.
54. Shew that the sum of the squares of the coefficients in
the expansion of (1 + a; + a:*)", where ti is a positive integer, is
2n
^* r r 2w2r*
55. Shew that, if n is any positive integer,
n(n^l) n{nl){n2){n3)
2(2r+l)"*" 2.4(2r+l)(2r + 3) "*"'"
= 9« r{r+l){r + 2)...{r + nl)
2r(2r+l)(2r + 2)...(2r + nl)*
v^.
CHAPTER XXIII.
ARTiAL Fractions. Indeterminate Coefficients.
295. In Chapter VIIL it was shewn how to express as
a single fraction the algebraic sum of any number of given
fractions. It is often necessary to perform the converse
operation, namely that of finding a number of fractions,
called partial fra^ctiom, whose denominators are of lower
dimensions than the denominator of a given fraction and
whose algebraic sum is equal to the given fraction.
296. We may always suppose that the numerator of
any fraction which is to be expressed in partial fractions
is of lower dimensions in some chosen letter than the
denominator. For, if this be not the case to begin with,
the numerator can be divided by the denominator until
the remainder is of lower dimensions : the given fraction
will then be expressed as the sum of an integral expression
and a fraction whose numerator is of lower dimensions
than its denominator.
297. Any fraction whose denominator is expressed
as the product of a number of diflferent factors of the
first degree can be reduced to a series of partial fractions
whose denominators are those factors of the first degree.
For let the denominator be the product of the n
factors a? — a, a? — 6, a? — c, ...; and let the numerator be
represented by F(x), where F{x) is any expression which
is not higher than the (n — l)th degree in x.
862 PARTIAL FRACTIONS.
We have to find values of A, B, C,... which are
independent of x and which will make
F(<») _ A B C ^
(a? — a)(a? — fe)(a? — c)... x — a a? — 6 x—c '
or, multiplying by (x — a)(x — b)(x — c) ,
F(x) = A{xb){xc) +jB(a?a)(a?c)
+ C(xa)(xb) (i).
In order that (i) may be an identity it is necessary
and su£Scient that the coefficients of like powers of x on
the two sides should be equal Now F(x) is of the
(n  l)th degree at most, and the terms on the right of (i)
are all of the (n — l)th degree ; hence, by equating the
coefficients of a?^ a?*,... a?*"* on the two sides of (i), we have
n equations which are sufficient to determine the n quan
tities A, B, G,
The values of A, B, 0,... can however be obtained
separately in the following manner. Since (i) is to be
true for all values of x, it must be true when x=ia'y and,
putting a? = a, we have F(a) = A(a — b){a'c) ; and
therefore A = F(a)/(a — b)(a — c) Similarly we have
B = F(b)/{ba)lbc)... ; and so for (7, D,....
We have thus found values of A, B, 0,... which make
the relation (i) true for the n values a, 6, c, ... of a? ; and
as the expressions on the two sides of (i) are of not higher
degree than the (n — l)th, it follows [Art. 91] that the
relation (i) is true for all valines of x.
Thus
F(a^ ^^ F{a) 1 ^
(it — a)(a? — 6)(a? — c)... (a — 6)(a — c)... x^a'
Ex. 1. Resolve , r . into partial fractions.
Assume r^±^— = ^ + ^ ;
then* Sx + 7=A{x2)+B(xl).
PARTIAL FRACTIONS. 363
In this identity put x=l; then 10=^. Now put x= 2; then
18 =B.
3« + 7 13 10
Thus
{xl)(x2) x2 x1'
Ex. 2. Resolve 1 ^} ,— ttt — r ^^^ partial fractions.
{xa){xb){xC)
Let (>c)(ca)(a> ),J_^J_ J^^
(a;a)(xo)(ajc) a;a xo ajc
then (6  c)(c  a) (a 6)=^ (xb)(xc) + B{x c) {x  a)
■\C{xa){xb),
Patting a; = a, we have (6 c){ca){ab)=A{a b) {ac); there
fore A=cb; and the values of B and C can be written down from
synunetry.
Thus (^<^)(<^"^)(<^^) <^^ I ^<^ , ^^
{x  a) («  6)(x c) oca x b x c'
Ex. 3. Resolve . . ... . — th^ — t, — ? into partial fractions.
a;(aj+l)(a; + 2)... (x+n)
Assume
1 :^_lJ1li. _L:4r_i lj^l
= ^+ —4+...+ ^ + ...+
a;(x + l)(x + 2) ... (« + «) x x+1 x + r x + n
Then, we have
l = i4o{(a; + l)(« + 2)...(x + n)}+i4i{x(x + 2)(x + 3)...(x + n)} + ...+
il,{x(x + l)...(x+rl)(x+r+l)...(x+n)}+...+^«{x(x + l)...(x + nl)}.
If we put x=0, all the terms on the right will vanish except the
first, and we shall have l=i4o x In, so that ^o=l/n.
To find the general term, put x= r; we then have
l=^^{(r)(r+l)...(l)(l)(2)...(nr)},
thatisl=(l)My[rnr; hence ^y=(l)»'/[rjnr.
Hence the required result is
lil
= T;:h;.. + (i)'7r^. :rr^+ Kl)
n 1
nr x+r x+n
x(x + l)...(x+n)"" n(x '"^^ '[
[See Art. 259. Ex.3.]
DX •4' OX \ f
Ex. 4. Express ; — ^ ,, ,. , ; in partial fractions.
(x  a) (x  6) (x  c) *^
pa^+qa+r
Ans, 2
(a— 6) (a  c) X  a *
364 PABTIAL FRACTIONS.
a^ + lB
Ex. 5. Besolve . rrr; — ^ into partial fraotions.
The fiaotors of a^+2x+5 are the complex expression8a;+l + 2t
and «+ 1  2t, where i is written for njl,
g»415 _ _A_ B C
^^®'*™® (a;l)(x«+2x + 5)=a;l '^'x+l + 2i'*' a;4l2t '
/. a;> + 16=il(a;+l+2i)(«+l2t)+B(xl)(* + l2i)
+ (7(«l)(«+l+2t>
Pat x=l; then 16=8^, so that A=2.
Puta:=l2t, then (l + 2i)2+16=B(22i)(4i), that ia
12+4t=B(8+8i); therefore B= 5^.. Change the sign of i
A — Jit
8i
in the valne of B, and we have C= 
Thus
2 + 2t*
a;»+16 2 34t 1 3t
(ajl)(a;2 + 2x + 5) xl 22iaj+l+2i 2 + 2i a; + l2i*
298. We have in the last example resolved the given
fraction into three partial fractions whose denominators
are all of the first degree, two of the fiictors of the denomi
nator being imaginary. Although it is for most purposes
necessary to do this, the reduction into partial fractions, of
a fraction whose denominator has imaginary factors, is often
left in a more incomplete state. Take, for example, the
fraction just considered, and assume
[It is to be noticed that we must now assume for the
numerator of the second fraction an expression containing
X but of lower degree than the denominator.]
Then a? + 15 = il (aj* + 2a? + 6) + (5a: + C)(a? 1).
Putting /c = 1, we have 16 = 8^, so that A = 2.
Put 4 = 2 in the above identity; then after transposi
tion a;*4aj + 5 = (5a? + (7)(a?l);
or, dividing by a?— 1, jBa? + (7 = — a? — 5.
a;'+15 2 a? + 5
Thus
(a?l)(a?*+2a?+5) xl a;' + 2a? + 5*
PARTIAL FRACTIONS. 865
299. We have hitherto supposed that the factors of
the denominator of the fraction which is to be expressed in
partial fractions, were all different from one another. The
method of procedure when this is not the case will be
seen from the following examples.
Ex. 1. Express , ttt^ ^. in partial fractions.
We may assume that
2x + 6 __ A B G D
(«l)8(j;3) = (a;l)3"*" (ajl)^"*" {xl)'^ xS'
or, olearing from fractions,
2x + 5=A[X'^3)+B{xl){xS)■\■C^ml)^xS) + D{xl)K
By equating the coefficients of x^^ x\ x^, x^ on the two sides of the
last equation, we shall have four equations to determine the four
quantities A, B, C, D, so that the assumption made is a legitimate
one. The actual values of A, By Cy D are not however generally
best found from the equations obtsJned by equating the coefficients of
the different powers of x. In the present case, the following method
is more expeditious.
Put x1 =y ; then we have
Now equate coefficients of y*, y\ y^^ y^, and we have 7= 2A;
2=:A2B; 0=B2C; andOsD + C
Whence A = l, 5=^^, C=^ ikndD=^ ,
iS 4 o o
„ 2a; + 6 11 7 11 11
Hence
(a;l)3(x3)~8(a?3) 2(a;l)8 4(ajl)' 8(a?l)'
(1 + «)*
Ex. 2. Express the fractional part of ^ __n\i ^^ partial fractions.
Assume
(l+a?)* A B C . X ,
(r:2^«= (Tr^+ (r2^+ (1^2^) + *^ "^*^^ expression.
Then
(1 + «)*= J + B (1  2a;) + C (1  2x)a + (1  2x)« X integral expression.
Now put 1  2«=y ; then
(l+a?)«=^ y= L(3*n3«iy+^^^^3'»2y2+ terms con
taining higher powers of y).
366 PARTIAJi FBACTIONS.
Also right Bide= A + By + Cy^+y^ x integral expression in y.
Hence, equating coefficients of y^, y\ y^, we have
3» n3*^ ^_ n(n 1)3'
lH2
l/J^ 800. The following examples will illustrate the use
^ of partial fractions.
Ex. 1. Find the coefficient of x* in the expansion of t^t^.a^
according to ascending powers of x.
w 1, 1 3 2
= 3{l + 3x + (3x)«+... + (3x)»+...}
2{l + 2x + (2x)2+... + (2x)*+...}.
Hence the required coefficient is 3**+^  2*'*"^.
Ex. 2. Find the coefficient of x^^*" in the expansion of /\ _ nl^s •
From Ex. 2, A^t. 299, we have
(l+x)* _3* 1 113**^ 1 w (n  1) 3*^ 1
(l2x)8~2»» (l2x)» 2* (l2x)2'*" 2*+i l2x
+ an integral expression of the (n  3)th degree. Whence the re
quired result.
Ex. 3. Shew that the sum of all the homogeneous products of n
dimensions of the three letters a, 6, c is equal to
a»Hi(c&)4 5*^ (g €) + €*+•> (&a)
(6 €)(€ a) (a  6)
The sum of all the homogeneous products of n dimensions is the
coefficient of x** in the product
(l+ax+aax3+...)(l + 6x + 6«x2+...)(l + cx + c2xa+...)[SeeArt.293];
that is in rz m — r—m n which will be found to be equal to
(1  ax) (1  bx) (1  ex)
a« 1 . 62 1 . c> 1
+ 77 TTT ; i V— +
{ab){ac) 1ax {bc)(ba) l6x^(ca)(c6) 1cx'
and the coefficients of x** in the expansions of these partial fractions
is easily seen to be
a»M 5*+9 ^»M
(a6)(ac)"^(6c)(6a)"'"(ca)(c6)*
which equals
a*+«(c5) + 6«^(ac)+c«>+^(5~a)
(6c)(ca)(a5)
INDETERMINATE COEFFICIENTS. 367
Ex. 4. To find the stun of all the homogeneous products of n
dimensions which can be formed from the r letters Oi* o^t at> * ^r*
As in the previous example, the sum required will be the co
efficient of a;* in rz rr^ rrz r , wluch wlU bc found
(1  tiix) (1  a^) (1  a^)
a,»^i 1
to be equivalent to Z. ^, r — = .
ajn+rl
Hence the required sum is S ; ^j^ — r — ; r .
801. Indeterminate coefficients. We shall con
clude this Chapter by giving two examples to illustrate a
method, called the method of indeterminate coefficients,
which depends upon the theorems established in Articles
91 and 281.
Ex. 1. Find the coefficient ota^ in the expansion, according to ascending
powers of x, of (1 + caf) (1 + c^x) (1 + c'x). . .(1 + c*«).
The continued product is of the nth degree in a; ; we may therefore
assume that
{l^cx)(l\e^x)...{l\6^x)=AQ+AiX+A^^{.,.+A,^+... + A^x^,
where Aq, A^ A^,... do not contain x.
Now change x into ex ; then, since Aq, A^, A^^ &c. do not contain
X, we have
(l + (^x)(l^c^x)...{l+&^^x) = Ao+AyCx+A^^x*+.„
+ Ar^i^aff^ + . . . + A^c^x\
Hence
= (1 + ex) (A^'\ A^ex + A^'^x^ + . . . + Ar4fai^ + . . . + A^e^x^),
Now equate the coefficients of d^ on the two sides of the last
identity, and we have
Ar + <^^Arx = ilrC*" + ^rl*^ J
" *•■" c**! «^l — ^ gr_i ^r\ \*/*
By continued application of (a) we have
g»r+l _ 1 c*""*^* — 1 c*"'*^' — 1
'*r='^ ;Fri ^'1="' ^^n • "'"' ^=rzr ^
_ (e^'  1) (<^^ l)...(c'  1) (c» 1)
e .C ...C,C (<,r_l)(crl_l)...(c>_l)(<,_l) ^^
368 INDETERMINATE COEFFICIENTS.
Ex. 2. To find the sum of the series V+^+d!^+ ...+nK
Let l*+2^+S^+ „.+n^=Ajn+Ain^+A^n^ (a)
for some particular value of n, where AnA^, A^ do iiot contain n.
The relation (a) will be true for n + 1 as well as for n, provided
13+23+32+. ..+n« + {n+l)a=Ji(7i+l)+Ja(n+l)2 + il8('» + l)';
or^ subtracting (a), provided
(n+l)2=^i + (2n + l)^a+{3n2+3n+l)i48.
Now the last relation will be true for all values of n if we give to
A^y A^,A^ the values which satisfy the equations found by equating
the coefficients of n\ n^ and n^^ namely, the equations
3i48=l, 8i<8 + 2^2=2, and A^^ A^{ A^^l^
&om which we obtain 6^^ = 2il2 = 3^8 ~ ^*
Hence, if the relation 12+22+...+n2=^n+s«*+5 n', be truQ
O 29 O
for any value of n, it will be true for the next greater value. But it
is obviously true when n=l; it will therefore be true when n=2;
and, being true when n=2, it must be true when n=3; and so on
indefinitely.
The sum of the cubes, or of any other integral powers, of the
first n integers can be found in a similar manner. [See also Art. 321.]
EXAMPLES XXIX.
Hesolve into partial fractions :
1 3a? x^\
3.
a^ + 7a: + 6* a;*5a; + 6'
(a;+l)(a;+3)(a; + 5) ** x{x + \y'
 8a; ^ aj' + oj+l
O. tt;: rsr;: ^ . 0.
(2  o;)* (1 + a;) * aj«4a;« + ic + 6*
„ a^3 . l + 7a;a:'
■ • Z ^v r Q TT • O.
(aj+2)(a:« + l)* ^' (1 +3a;/(l  10a;)*
aj'ai + l 5 9a;
y» ■; — a , V , TT^ . 10.
(a;»+l)(ajl)«' (l3a;)«(l +aj) '
X
EXAMPLES. 369
11 6a;* + a;  1 »' + 2
(rB«+l)(a:~2)(a: + 3)" (aj  2)* (a;« + 1) '
,„ a;' + a; . a;'a;+l
(a;l)^(ajV4)* " (aj l)«((B2)(a;'+ I)"
Vl5. ,. Lt??^ ,.. 16. ^^^^
a«((B+2)» (»+!)• • a;«(a; + 2)'(a;l)
17. Find the coefficient of a;" in the expansion of
a; + 4
a;* + 5a: + 6 '
18. Find the coefficient of x* in the expansion of
a;2
(a;+2)(ajl)'*
19. Shew that the coefficient of «*""* in the expansion of
a; + 5 . , 1
isl
(a'l)(a;+2)
tSn*
20. Find the sum of the n first coefficients in the expansion
. 3 2a;
l2a;3a:»*
21. Find the sum of the n first coefficients in the expansion
. 25a;
(l5a:)(l3a;)(l2aj)*
(1 + a;)"
22. Find the coefficient of a:" in the expansion of y^ ^ .
Find also the sum of the n first coefficients.
23. Shew that the coefficient of a;'*'*''' in the expansion of
s. A. 24
870 £XAMPL£S.
24. Shew that
25. Shew that the coefficient of 2;""* in the expansion of
{(1  sr) (1  cz) (1  c'z) (1  c««)}^ is
(1  C) (1  c"+^) (1  c^^)/(l  c) (1  c') (1  c").
26. Prove that
a(bc){bcaa')(a'^'a"^) b {c  a) (ca  hb') jb'^  h'"")
a'a' ■*" b'b'
c(ab) (ab  cc') (c"  Q
:=_ ^ (5 . c) (c  a) (a  5) (6c  aa') {ca  56') (a6  cc') ZT^.,,
where aa' = 66' = cc', and ff_ , is the sum of the homos^eneous
products of a, bj c, a, 6', c' of m — 3 dimensions.
27. Shew that the product of any r consecutive terms of
the series 1c, 1—0% l—c%,.. is divisible by the first r of
them.
28. Shew that, if c be numerically less than unity,
(1 + caj) (1 + c^x) (1 + c'^x) ...to infinity
1 ^ c« c^^^+^> „
~^ + lc'*''^(lc)(lc^)^'^'*"(lc)...(lc")^ ^•••
29. Shew that, if c be numerically less than unity,
(1 + co;) (1 + (^x) (1 + c'a?) ... to infinity
1 c c* ^ c^ ^
30. Shew that, if c be less than unity,
1 ^ X a?
(laj)(lca;)(lc«a;)... ^1c (lc)(lc»)
a?
^(lc)(lc*)(lc«) ■*■••• [^^^ss.]
EXAMPLES. 371
3L Shew that, if c be less than unity,
{l+cx)(l + c'x){l+c'x),„ _, . l+c (l + c)(l+c') .
(la:)(lca;)(lc'a:)... "^'*'lc^ (lc)(lc*)
[Gauss.]
32. Shew that the coefficient of a;' in the expansion of
(1 + cx)(l + <^x)(l f c'a?) ...
(l«»)(lc*ic)(lc'a;)...
(l + l)(l+o)...(l+c)
(lc)(lc")...(lcO '
c being less than unity.
33. Shew that
+ 5 +  a I ^ 5 — t • ••
105 1003 1— a'x 1  a^x
\ X a? ix?
"^ 1 ZT. "^ T 157. "*" 1 TsT. "•■ ••"
1y 1ay 1a'y loTy
34. Shew that
a; 2aj' Saj* 4aj*
1a; laj* laj» 1a;
03 a;^ 05*
A* . .•
" (1 «)• ^ (1 »»)• ■*" (1 ic»)''^ •••
35. Shew that Lambert's series, namely,
X a? x^ 03*
is equivalent to
X z + X* = 5 + ar :i » + ... [Clausea J
1a; las* I03' *• '
24—2
CHAPTER XXIV.
Exponential Theobem. Logarithms. Logarithmic
Series.
302. The Exponential Theorem. If 1/n be nu
merically less than unity, ( 1 +  ) can be expanded by
the Binomial Theorem ; and we have
\ nj n 1.2 w'
nx{ruc — l)(nx—2) 1 nx(nx—l)...{nx — r{l) 1
.2.3 TT \r n
which may be written
(l + j =1+0. + — __+ — ^ —
X (x J... (a? )
. V nJ \ n J ^
Putting a? = 1, we have
. ^ _ i_! filVi?)
, ('S4^) ,
+...
EXPONENTIAL THEOREM. 373
xlo) I x\x )(a? — J
,.1. n , \ nj\ nl ^ Y
^^^'^^^^ 1^ + •••} •
The above relation is true for all values of n however
great, and therefore when n is infinite; but when n is
infinite, Ijn is zero, and the relation becomes*
Denoting the series 1 + 1 + t^ + j^ + ... + i + ... by e,
we have the Exponential Theorem, namely
«. ^ (Xj of
It should be remarked that the above series for e* is
convergent for all values of x [Art. 278].
303. The quantity e is of very great importance
in mathematics.
It is obvious that it is greater than 2 and it is clearly less
than 1 + 1 + 2"* + 2"' + 2"* + . . ., and therefore less than 3.
Its actual value can be found to be 2*71828....
* This requires more careful examination not only to find the limit of
each term, but also because the limit of a sum is not necessarily equal
to the sum of the limits of its terms unless the number of the terms
Infinite, This examination is however omitted here for the inyestigation
in Art. 304 is preferable.
874 EXPONENTIAL THEOREM.
To prove that e is an incommensurable number.
If possible, let e = m/n, where m and n are integers ;
then we should have
^=1+1 + 1^+ + g+j^+^+l^ + 
Multiply both sides by In; then all the terms will
become integral except
7i+l^(7i + 2)(n + l)^(n + 3)(/i + 2)(M + l)^
Hence
n + 1 ' (n + 2)(7i+l) (w+3)(7H2)(tt+l)
must be equal to an integer; but this sum is less than
111
+ y ^ + p =^ + . . ., and therefore less than
w + 1 (n + iy (n+ 1)
/ 1 1 . ) that is less than  . But an integer
n + l/Vw + l/ n °
cannot be less than 1/n ; it therefore follows that e cannot
be equal to the commensurable number m/n,
304. The following proof of the Exponential Theorem
is due to Prof. Hill*. It will be seen that it only assumes
the truth of the Binomial Theorem for a positive integral
exponent.
S r
Let /(to) denote the series l+m + Tzr+ + 7+....
2 r
Thus/(m)=l+m + ^+ + ^+ ,
* Proceedingt of the Cambridge Philosophical Society, Vol. v. p. 415.
SubstontiaUy tne same proof is however given in Cauohy's Analyse Algi
brique.
EXPONENTIAL THEOREM. 375
f(n)^l+n + ^ + + 1 +
and/(m+n) = l+(m + ?i)+ ^ Z ^ +...+ ^ 7 ^ +...
Now the coefficient of mV in/(m) ^f(n) is i — r ;
\r [«
and in/(m + n) the term mV can only occur in 
r + 8
r + « 1
and its coefficient will therefore be y=j= , that is
jr 5 r\'8
1
FY'
Hence, as the series /(m), f{n) and /(m + ri) are
convergent for all values of m and n, and the coeffi
cient of any term mV is the same in/(m) xf(n) as in
f(m + n), it follows from Art. 280 that
f{m)x/(n)=f(m + n) (i)
for all values of m and tu
Now let a? be a positive integer; then from (i) we
have
/(I) x/(l) x/(l) + to X factors,
=/(l + 1 + 1 + to X terms),
.. {/(l)r=/(^) (ii).
Next let a? be a positive fraction  , where p and q are
positive integers. Then from (i)
[^ (l)F=^(f ^f ^f + "^ « *^H =^<^>
{/(l)}',from(ii);
Hence, for all positive values of x, {/(I)}'*' ='/{^)»
376 EXPONENTIAL THEOREM.
Lastly, let x be negative, and equal to — y, so that y is
positive ; then/( y) xf(y) =/(0) from (i) ; but/(0) = 1,
therefore /( y) = l/f(y).
Hence
1 1
/(^) =/( y) = jTjT = i f/i\\v > since y is positive,
={/(i)r={/(i)r.
Hence, whatever a? may be,
{/(i)r =/(«')•
But /(i) = i+^+l + l + = ,,
therefore e'=/(a;) = 1 +^ + ^' + + f+
305. To shew that
n(nl) + ?L^:il>(n2)"... = n.
We have from Art. 304
Also, by the binomial theorem,
a;+~+^+...j is zero, if r is
less than n, and is 1 if r=n.
Also the coefficient of x^ in e"*n«<'»i)«+^^^—^ «(««« ... is
ij«,«(nl)r+!L^(„2,._...j.
Hence, equating the coefficients of a?* in the expansions of the
two expressions for (e« 1)«, we have
i n»n(nl) + ?i^)(„_2)» ...1=1.
n"
n
EXPONENTIAL THEOREM. 377
The above theorem may be generalised as follows:
We have
and
Hence, equating coefficients of x^ in the two expressions for
(go* _ gftxj* ^e have
If we put na=:x and ba=yt the last result becomes
a;«n(a;+y)"+^^^'(x + 2y) ... = (!)" .y" n.
We have also, if k be any positive integer less than n^
x*n(a; + y)*+ **^!*~ ' (a?+2y)* ... ton+1 terms=0.
X a^
The following particular cases are of importance, k being less
than n.
1*_ n2*+^^^^^ 3* "... to w + 1 terms=0,
X • ^
and m*w(ml)*+— ^^^(m2)* ... ton+1 terms=0.
X • A
EXAMPLES XXX,
Ex. 1. Shew that the limit when n is infinite of f 1 +  ] is ^.
(\ *
1 + ) is ^ .
Ex. 8. Shew that
n**in(nl)»+i+^^^^«2)**i« ... =lnn + l.
Ex. 4. Shew that
378 LOGARITHMS.
Ex. 6. Shew that
2 4 6
Ex. 6. Shew that e^=j^+ _ + _+...
3 [6 7
Ex. 7. Shew that
3 1 + 2 . 1 + 2 + 3 . 1 + 2 + 8 + 4 .
2^" "^ [2 "^ [¥""*■ g '^•••
Ex. 8. Shew that
/.111 \^ ^ A 1 1 1 V
V^ + 2+[4 + [6+j=l + (l + J3 + ^+[7+j •
Ex. 9. Shew that
_i_ J_ 1 1
* ""1.3"*'l.2.3.5'*''^1.2.3...(2nl)(2n + l)"^*
Ex. 10. Shew that
e
+ l^+4+l6 + il+l3+^+.j
Ex. 11. Shew that
e«+l L 1 1 J^ > h 1 Jlj.Jl I
Ex. 12. Shew that the coefficient of x* in the expansion of
(l + 2x) (l + 2rc)« ( l + 2x)3 2»6
Logarithms.
306. Definition. The index of the power to which
one number must be raised to produce a second number is
called the logarithm of the second number with respect to
the first as base. Thus, if a' = y, then x is called the
logarithm of ^ to the base a, and this is expressed by the
notation x = log^ y.
PROPERTIES OP LOGARITHMS. 379
We proceed to investigate the fundamental properties
of logarithms, and to shew how logarithms can be found,
and how they can be employed to shorten certain approxi
mate calculations.
307. Properties of Logarithms. The following are
the fundamental properties of logarithms.
I. Since a® = l, for all values of a, it follows that
log, 1=0.
Thus the logarithm, of 1 is 0, whatever the base may
be.
II. Iflog„a? = a, loga2^ = /3, log„^ = 7, ...
then a? = a*, y = a^^ ^ = aY,...;
.'. ayz ... =a*.a^.ay... = a"+^''"Y+ •
= log. « + logay + log„ ^ + ...
Thus the logarithm of a product is the sum of the
logarithms of its fojctors.
III. If log. a: = a, and log. y = ^\
then a? = a", y = a^, and .*. aj^y = a*"^;
.. log. (a? ^ y) = a  /3 = log.a?  log. y.
Thus the logarithm of a quotient is the algebraic differ
ence of the logarithms of the dividend and the divisor.
IV. If a? = a* ; then af^ = a"*, for all values of m.
Hence log. a?*" = ma = m log. x.
Thus iJie logarithm of any power of a number is the
product of the logarithm of that number by the index of the
power.
V. Let log. x = a, and log,, a? = /8 ; then a; = a* = &^ ;
and hence a = 6*, and a^ = b.
380 LOQABITHMIC SERIES.
Therefore  = log^a, and 5 = logj).
Hence log„ b x log^ a = ~ x  = 1.
Also )8 = a log„ a, that is log^, a? = log. x . log„a.
Hence ^Ae logarithm of any nwrnher to the base b mil be
foimd by multiplying the logarithm of that number to the
base a by the constant multiplier logt, a,
308. The logarithmic series. Let a = e*, so that
fc = log, a; then a* = e** = e*^*««» Hence from Art. 304,
we have
Put a = 1 + y ; then we have
(l+yy^l + xlog^(l+y) + hxlog,(l^y)Y^...
Now, provided y be numerically less tha/n imitv, (1 + y)*
can be expanded by the binomial theorem ; we tnen have
= l+a?log,(l + y)+2{^log,(l + y)P+...
The series on the right is convergent for all values of x
and y, and the series on the left is convergent for all values
of X provided y is numerically less than unity. Hence, for
such values of y, we may equate the coefficients of x
on the two sides of the equation. We thus obtain
iog.(i+y)=y^+^ +(ir^+...
This is called the logarithmic series.
LOGARITHMIC SERIBS. 381
Ex. 1. To express a*+ &** in terms of powers of ab and a + 6.
From the identity (1  ox) (1  6«) = 1  (a + 6) a; + abx^
where t is put for a+ 6 and p for ab, we have
log, (1  ax) + lege (1  bx)=logg (1  8x +>a;2).
Hence ( <»^ + 2 + "3" + ) + ( fr^+g +3 + J
Equate the coefficients of x^ on the two sides of the last equation.
[This is allowable since the series can clearly be made convergent by
taking x sufficiently small.] Then the coefficient of «* on the left is
(a*+l>*). On the right we have to pick out the coefficient of a;*
from the terms (beginning at the highest in which it can appear)
^ (« pxy + ^j [8 px)''^ +^32 ^* "1^^)"'*+ »
the coefficient of x* is therefore
1^+ JL {_(„_!) ..«^j + J_ l {n2)(nZ) 1
n^ n1* ' "^' »2 I 1.2 "^ J
Hence we have
o* + 6* = (a + &)*  na6 (a + 6)*2 + ^^"7/^ a^t^ (« + 6)«*
... + (!)."("' ^)(''^2)<"^'+^)a.ftr(, + 6).a,+
Ex. 2. Shew that, if a + & + c = ; then will
10(o7 + 67+c7) = 7(aa+6« + c2)(a8 + 6»+c6).
Put p for bc+ca+abt and 9 for a&c ; then we have the identity
(1  ax) (1  bx) (1  cx)as 1 jpx'  gx*.
Now take logarithms, and equate the coefficients of the different
powers of X in the two expansions. This gives  (or + fr*" + C) in
terms of p and q, and the required result follows at once. [See also
Art. 129.]
382 cauchy's theorem.
Ex. 8. To express o'^+t^+c* in tenns of ode and bc + ca+db, when
a + 5+c=0.
Put p = bc+ca¥dbf and q=abc; then we have the identity
(1  cue) (1  bx) (1  ex) = 1 pa^  qx^.
Hence, by taking logarithms, and equating the coefficients of like
powers of a;, we have
 (a*+ 6* + c**) = coefficient of a;* in S  a;^(? + qx)^,
n T
which gives the required result.
The terms ia^x^{p + qx)*" which contain afi^*^ are
T
+ . . . + ^ — 3j a;®»»*~2 (jp + ^x)*"*! + — x«»» {p + ga;)""*.
Now by inspection we see that the coefficient of x^^^ in each
of the above teams in which it occurs contains jpg as a factor ; and
also that the coefficient of x^"*^^ in each of the terms in which it
occurs contains p!^q as a factor.
Hence, when a + 6 + c=0, a*+6*+c* is algebraically divisible by
abc{bc+ca+ab) when n is of the form 6»»l, and a^+b'^{'C'^ is
algebraically divisible by abc{bc + ca+aby^ when n is of the form
6m+l.
If we put c=(a+6), bc+ca+ab becomes (a^+oft+ft*), and
we have Cauchy*s Theorem, namely that a*+6* (a+i)** is divisible
by ab{a+b) [a^+ab + b^) when n is of the form 6wil, and by
ab {a + 6) (a^ + a6 + 6')' when n is of the form 6m + 1.
[See papers on Gauchy*s Theorem by Mr J. W. L. Glaisher and
Mr T. Muir in the Quarterly Journal^ Vol. xvi., and in the Messenger
of Mathematics, Vol. vin.]
809. In order to diminish the labour of finding the
approximate value of the loffarithm of any number, more
rapidly convei^g series are obtained fbc>in the funda
mental logarithmic series.
Changing the sign of ^ in the logarithmic series
log.(i+y) = y^ + f^ + (i).
we have
log.(iy)=yff' (ii).
LOGARITHMIC SERIES. S83
1 + V
Hence log, j— ^ = log. (1 + y)  log, (1  y)
= 2(y+ +  + ) ....,.(iii).
Put — for  — " , and therefore for y ; then
n 1 — y m + w *^
We are now able to calculate logarithms to base e
without much labour. For example : —
Put m = 2, w = 1, in formula (iv) ; then
log.2 = 2
"l" o • «Jl I !• • rk5 "T" • • •
3* 3« • 6* 3'
from which it is easy to obtain the value log, 2 = '693147...
Having found log, 2, we have from (iv)
log,3log,2 = 2^ + J.i+J.^+...J = 405465....
Hence log, 3 = '6931 4*7 + '405465 = r09861.
Proceeding in this way, the logarithm to base e of
any number can be found to any requisite degree of
approximation.
310. Logarithms to base e are called Napierian or
natural logarithms.
The logarithms used in all theoretical investigations
are Napierian logarithms ; but when approximate numeri
cal calculations are made by means of logarithms, the
logarithms used are always those to base 10, for reasons
wEich wiU shortly appe/r: on this account logarithms
to base 10 are called Cormnon logarithms.
We have shewn how logarithms to base e can be found;
and having found logarithms to base 6, the logarithms to
base 10 are obtained by multiplying by the constant
factor logj^e, or by l/log,10. [Art. 307, V.] This constant
factor is called the Modulus, its value is •43429...
384 EXAMPLES.
EXAMPLES XXXL
1. Shew that log (x + n) — log x + log (! + ] +
2. Shew that log, 7l2 = l + (^ + 3) j + (j + 5)51 +
{hlji^^Cs^Di" ^o infinity.
11 l^^ 1^
3 9 "^ 5 9* "^ 7 9'
1 11 11 11
3. Shew that log, ^10 = jl + ^^ + ^;^+ i:^+ to
infinity + {9 + 3 98+ 5 95 + y 97+ ••— to infinityj .
4. Shewthatlog,2i=^3f3l^ + ^+
to infinity.
5. Shew that ^A_+^^ + ^+ to in
finity = 3 log, 21.
6. Shewthatlog,^^ = 2{2^+1^2^
1 1
5(2a;l)
~^^ }
8. Shew that
a + x 2ax 1 / 2ax V 1 / 2ax \*
a + x _ 2ax 1 / 2ax y 1 / 2ax V
+
EXAMPLEa 385
9. Shew that
2x 1 / 2a; \» 1 / 2x
^/ x' x'' \ 2x 1 / 2a; \» 1 / 2a; \*
10. Shew that
{iog,{i + x)Y =2 ^<^l(^+iy +1(1 +l+iy* ...].
11. Shew that, if log^ (1 + a; + a;*) be expanded in powers
1 2
of X, the coefficient of a;" is either  or — , and distinguish the
cases.
12. If loga (1  a; + a;*) be expanded in ascending powers of
X in the form a^x + a^x' + a^af + , then will a^^a^ + a^^ .,,
= I log. 2.
]^ 1. /]j ^ />j^
13. Expand log^ = ^ in ascending powers of x.
A "" a; + a;
14. Shew that
1 a; a;' x^
n w (w + 1) ri (t* + 1) (/i + 2) w (w + 1) (t* + 2) (/* + 3)
rl X x' a;° I
*^tw l(w+l)'*' 2(w + 2) [3(n + 3)'*' /•
15. From the identity 2 log (1 — a;) = log (1 — 2a; + a;'), prove
that2n.2' + "(^^> 2 "<^t^^:^> 2'V...=2.
1 . J 1 . J . o
16. If logg = 3 5 be expanded in a series of positive
L ^ X "~ a* T" a;
1 3
integral powers of x, the coefficient of x" will be  or  accord
ing as n is odd or even.
S.A. 25
386 EXAMPLES.
17. Shew that the coefficient of of in the expansion of
e"* — 1 1
= J is7{r + 2'' + 3'+ +w'}. Hence find the sum of n
terms of the series 1" + 2* + 3* + ..., and also of 1* + 2* + 3' + ...
18. Shew that, if a^ be the coefficient of a?' in the ex
pansion of e^, then
_ 1 rr 2;; 3^ I
Hence shew that
18 Q8 OS
[T"^2*j3'^"^'''
and that
19. Shew that
f n n(nl) n(n^l)(n2) \
\ V V.2' ■*■ 1^2«.3« j
= 1 +
^^. n. (^ + l)(^+2) . (n + l)(n + 2)(n + 3) ^
\nf L) \ p ^5 "*" 1* 2' 3* T" •••
20. Shew that the sum of w terms of the series j + o + o + • • •>
beginning at the {n + l)th, becomes equal to log, 2 when n is
increased without limit.
21. Shew that
log.(l+n)<Y+H+ ... +<1 +log,(l+n).
22. Prove the following : —
(i) {x + yY x^y* = Ixy {x + y) {a? + xy •\r y*)*,
(ii) {x + 2/)"  a;"  y'' = l\xy ix v y) {od" + xy + y')
{(a^ + xy\ ^Y + a:*/ (aj + yf],
(iii) {x + y)***  a"  y ■ = 1 3a^ {x + 2/) (a* + ary + 2^')'
{(a;" + a^ + 3^")^ + ^afy" (» + y)'}.
ittt
COMMON LOGABITHMS. 387
23. Shew that a^ + y^^(x + yf
o « / o\ «8^ n(nZ)(n — 4) (n  5) ,_« .
3.4
71 (7 irl)...(7^3r+l)
"" 374:::2;; ^ ^ ■*■•••'
where p = ix? + xy + y^ and g' = o^ (a; + y).
24. Shew that, (i) if w be any uneven integer, (6 — c)" +
(c  a]" + (a  6)" wiU be divisible by (6  cf + (c  a)» + {a  6)»;
(ii) if w be of the form 6m ± 1, it will be also divisible by
(6c)» + (ca)' + (a6)*; and (iii) if w be of the form
6m + 1 it will be divisible by (6  c)* + (c  a)* + (a  6)*.
Common Logarithms.
311. In what follows the logarithms must always be
supposed to be common logarithms, and the base, 10, need
not be written.
If two numbers have the same figures, and therefore
differ only in the position of the decimal point, the one
must be the product of the other and some integral power
of 10, and hence from Art. 307, II. the logarithms of the
numbers will differ by an integer.
Thus log 4215 = log 4215 + log 100 = 2 + log 4*215.
Again, knowing that log 2 = '30103, we have log '02
= log (2 r 100) = log 2  log 100 = 30103  2.
On account of the above property, common logarithms
are always written with the decimal part positive, _ Thus
log 02 is not written in the form  169897 but 230103,
the minus sign referring only to the integral portion of
the logarithm and being written above the figure to which
it refers.
Definition. When a logarithm is so written that its
decimal part is positive, the decimal part of the logarithm
is called the mantissa and the integral part the character
istic.
25—2
388 CHARACTEKISTICS FOUND BT INSPECTION.
312. The characteristic of the logarithm of any number
can he written down by inspection. For, if the number be
greater than 1, and n be the number of figures in its
integral part, the number is clearly less than 10* but not
less than 10*'\
Hence its logarithm is between n and n — 1 : the
logarithm is therefore equal to n— 1 +a decimal.
Thus the characteristic of the logarithm, of any number
greater than unity is one less iha/n the nv/mier of figures in
its integral part.
Next, let the number be less than unity.
Express the number as a decimal, and let n be the
number of ciphers before its first significant figure.
Then the number is greater than 10"*"* and less than
10"*.
Hence, as the decimal part of the logarithm must be
positive, the logarithm of the number will be — (w + 1) +
a decimal fraction, the characteristic being —(n + 1).
Thus, if a number less than unity be eapressed as a
decimal, the characteristic of its logarithm is negative and
one more than the number of ciphers before the first signifi
ca/nt figure.
For example, the oharaoteristio of the logarithm of 8571*4 is 8,
and that of 00035714 is 4.
Conversely, if we know the characteristic of the
logarithm of any number whose digits form a certain
sequence of figures we know at once where to place the
decimal point.
For example, knowing that the logarithm of a number whose
digits form the sequence 85714 is 8*55283, we know that the nnmber
must be 3571*4.
313. Tables are published which give the logarithms
of all numbers from 1 to 99999 calculated to seven places
of decimals : these are called ' sevenfigure ' logarithms.
For many purposes it is however sufficient to use five
figure logarithms.
J
USE OF TABLES OP LOGARITHMS. 389
In all Tables of logarithms the mantissae only are
given, for the characteristics can always, as we have seen,
be written down by inspection.
In making use of Tables of logarithms we have, I. to
find the logarithm of a given number, and II. to find the
number which has a given logarithm.
I. To find the logarithm of a given number.
If the number have no more than five significant
figures, its logarithm will be given in the tables. But, if
the number have more significant figures than are given
in the tables, use must be made of the principle that
when the difference of two numbers is small compared
with either of them, the difference of the numbers is ap
proximately proportional to the difference of their loga
rithms. This follows at once from Art. 308, for
\og,,{N^x)^\og,,N^\og,,{l +.) = /.log.(l +)
/ OC Su \ CD OC
= /A (;^""i;^+) "^ ^^'W approximately, when ^ is
small, /L6 being the modulus 1/log, 10.
An example will shew how the above principle, called
the Principle of Proportional Differences, is utilised.
Ex. To find the logarithm of 357*247.
We find from the tables that log 35724= 5529601, and log 85725
= 5529722; and the difference of these logarithms is '0000121.
Now the difference between 8*57247 and 3*5724 is ^ths, of the
difference between 35724 and 3*5725 ; and hence if we add ^hs. of
0000121 to the logarithm of 85724 we shall obtain the approximate
logarithm of 857247. Now ^ths. of 0000121 is 00000847, which
is nearer to 0000085 than to 0000084. Hence the nearest approxi
mation we can find to the logarithm of 857247 is 5529601 + 0000085
= 5529686.
The characteristic of the logarithm of 857*247 is obviously 2, and
therefore the logarithm reqnir^ is 25529686.
II. To fi/nd the number which has a given logarithm,
"Pot example, let the given logarithm be 4*5529652.
We find from the tables that log 85724 = 5529601 and that
log 35725= 5529722, the mantissa of the given logarithm £alling
890 COMPOUND INTEREST AND ANNUITIES.
btiween tbeie two. Now the difference between 5529601 and the
giren logarithm 10 r^ of the difference between the logarithniB of
8'6734 and 8*5725; and hence, by the principle of proportionai
differencei, the number whose logarithm is *5529652 is
8*5724+ ~ X 0001 = 85724 +*00004= 357244.
[The approximation could only be relied upon for one figure.]
Thui *5529652srlog 8*57244, and therefore
4'5529652=log 000357244.
Compound Interest and Annuities.
814. Tho approximate calculation of Compound In
teroBt for a long period, and also of the value of an annuity,
can be readily made by means of logarithms.
All problems of this kind depend upon the three fol
lowing : — [The student is supposed to be acquainted with
the arithmetical treatment of these subjects.]
I. To fi/nd the a/mount of a given sum at compound
interest, in a given number of years and at a given rate
per cent, per a/rmwra.
Let P denote the principal, n the number of years,
lOOr the rate per cent, per annum, and A the required
amount.
Then the interest of P for one year will be Pr, and
therefore the amount of principal and interest at the end
of the first year will be P (1 + r). This last sum is the
capital on which interest is to be paid for the second
year ; and therefore the amount at the end of the second
year will be {P (1 + r)} (1 + r) = P (1 + r)\ Similarly the
amount at the end of n years will be P (1 + r)".
Thus J. = P (1 + r)* ; and hence
log A =logP + nlog(l + r).
If the interest is paid, and capitalised, half yearly, it
can be easily seen that the amount will be P f 1 + ^ j .
ANNUITIES. 391
Ex. Find the amount of £360 in 26 years at 6 per cent, per annum.
Here P=350, »'=Jqq and n=25;
.\ log ^= log 360 +26 log (l + iQ^)
= log 360 + 26 (log 106  log 100).
From the tables we find that log 360=26440680 and log 106 =
20211893; hence log ^=30738006. Whence it is found from the
tables that ^=£118622.
II. To find the present value of a sum of money which
is to he paid at the end of a given time.
Let A be the sum payable at the end of n years, and
let P be its present worth, the interest on money being
supposed to be lOOr per cent, per annum. Then the
amount of P in n years at lOOr per cent, per annum
must be just equal to A.
Hence from I. P = ^ (1 + ry\
in. To find the present value of a/n annuity of £A
payable at the end of each of n su^ccessive years.
If the interest on money be supposed to be lOOr per
cent, per annum ; then from II.
The present value of the first payment is ^ (1 + r)"*
second 4 (1 + r)"^
nth u4(l+r)"~.
Hence the present value of the whole is
A\
1 1
=^n
l+r"*"(l+r)''^''"(l+r)J~r r 0 + r)T
Ex. Find the present value of an annuity of £30 to be paid for 20
years, reckoning interest at 4 per cent.
Here 4=30, «=20.r=j^ = 2^g.
392
EXAMPLES
Henoe the present value = 80x25i(hb) \ •
Now log ^)**=20{log26log26}
= 20 {13979400  1 4149733}
=20 (0170333)=  840666 =1669334
=log 456889, from the Tables.
Henoe the yalae reqmred=30 x 25 x (1  456889) =£4077,..
EXAMPLES XXXIL
The following logarithms
log 102
log 1025 =
log 1033
log 104
log 105
log 1 06
log 11467 = 0594498
log 11468 = 0594877
log 12258 = 0884196
log 12620 = 1010594
log 14816 = 1707310
log 14817 = 1707603
■0086002
•0107239
0141003
0170333
•0211893
•0253059
1. Find ^105.
are given
log 16386 .
log 16387 =
log 17292 =
log 17349:
log 2
log 20829 :
log 3
log 30832 =
log 44230 :
log 51
log 5577 :
log 63862 ■•
log 74297 ■■
log 74298 :
2. FindjySl.
•2144730
•2144995
•2378452
•2392744
3010300
3186684
•4771213
4890017
•6457169
•7075702
•7464006
•8052425
•8709713
•8709771
3. Find the amount of £100 in 50 years at 5 per cent,
per annum.
4. Shew that money will more than double itself in 15
years at 5 per cent, per annum, and in 18 years at 4 per cent,
per annum.
EXAMPLES. 393
6. Find the amount of j£500 in 10 years, interest at 4 per
cent, being paid half yearly.
6. The number of births in a certain country every year
is 85 per 1000 and the number of deaths 52 per 1000 of the
population at the beginning of every year: shew that the popu
lation will be more than doubled in 22 years.
7. A man invests £30 a year in a Savings Bank which
pays 2^ per cent, per annum on all deposits. What will be
the total amount at the end of 20 years?
8. What sum should be paid for an annuity of ^100 a
year to be paid for 40 years, money being supposed to be worth
4 per cent, per annum 1
9. A corporation borrows £30000 which is to be repaid
by 30 equal yearly payments. How much will have to be paid
each year, money being supposed to be worth 4 per cent, per
annum)
10. A house which is really worth £70 a year is let on a
lease for 40 years at a rent of £10 a year, the lease being re
newable at the end of every 14 years on payment of a fine.
Calculate the amount of the fine, reckoning interest at 6 per
cent.
CHAPTER XXV.
Summation of Series.
315. We have already considered some important
classes of series, namely the Progressions [Chapter xvn].
Binomial series [Art. 288], and Exponential and Logarith
mic series [Chapter xxiv]. In the present chapter some
other important types of series will be considered.
316. The nth term of a series will be denoted by u^,
and the sum of n terms by S^. When the series is con
vergent its sum to infinity will be denoted by 8^.
317. No general method can be riven by which the
summation of series can be eflFected; but in a great
number of cases the result can be obtained by expressing
the general term of the series, u^, as the diflference of
two expressions one of which involves n — 1 in the same
manner as the other involves n.
For example, in the series
a a a
a!{x + a) '^(x\a)(x+ 2a) (a? + 2a) (x + 3a) *"'
the nth term, namely == , is equal to
(x\n— l.a)(x+na)
 7 vv ; • Hence the series may be written
a? + (n — 1) a a? + na •^
SUMMATION OF SERIES. 395
/I _ Ji\ / 1 1 \ / 1 1 \
\a) x\a) \x+a a+2a/ \x\2a x + SaJ
+ ■! — —p Tv : [ ; and it is now obvious that all
[x +{n — l)a x{na)
the terms cancel except the first and last ;
i_ „ 1 1 na
hence &. = —
X X \ na x{x + na) '
Ex. 1. Find the Bum of n terms of the series
_L _L J^ ^
1.2 "^2. 3 "•"3.4'*' '^n(n + l)'*""
Ans, 1
w+1*
Ex. 2. Find the sum of n terms of the series
2'*'3'*" li"*" "^ jn + l "^
Here u^=. 7 = . Ans. 1 — r .
" \n \n+l \n + l
Ex. 3. Find the snm to infinity of the series
111 1
3i 42 63 (n+2)^
Here tL.= —^ x . Ans. jr.
^ n + l n + 2 2
Ex. 4. Find the snm to infinity of the series
3 6 7 2n+l
Mi + ;io—r, + +oT^Tr,+
12. 2« ' 22. 3« ' 32.42 n2(n+l)»
Here u^=s  7 — tto • ^rw. 1 .
" n^ (n+l)2
Ex. 5. Find the snm of n terms of the series
1 2 3 n
1.3 "^1.3.6'*' 1.8. 6. 7"*" "*"l.3.6...(2n+l)'
fo^ L_ 1 "I
L * 1.3.5...(2nl) 1.3.6...(2»l)(2n+l)J'
"*""• 2 r"l.3.5...(2n+l)[
396 SUMMATION OP SERIES.
Ex. 6. Bum to infinity the series
1.3 3"*'8.6 • ^^B.T 33"*" ■^(2ul)(2ii + l)3»'^
r«. n+l If 3 1 \
L ^^ (2n 1) (2n+l) " 4 V2n 1 2;m3J *
^*^2^:ri3ii=i2^:ja3^J ^**' 4
Ex. 7. Find the stun to infinity of the series
1111 1
{■ Aru, .
2a_i^4a1^6a1^8*^l^ 2
Ex. 8. Find the sum of n terms of the series
+ ..
1 1
4n«.
{lx)^ (lar)(la;»+i)*
318. To find the sum ofn terms of the series
{a(a + b) .. .(aH r1.6)} + {(a + 6) (a + 26). .. (a + rb)}
+ ... + {(a + n 1 ,b)(a+nb) ...(a + n + r — 2 .6)} + ...
In the above series (i) each term contains r factors,
(ii) the factors of any term are in arithmetical progression,
and (iii) the first factors of the successive terms form the
same A.P. as the successive factors of the first term.
Consider the series which is formed according to the
same law but with one factor added at the end of every
term, and let v^ be the nth term of this new series, so that
t;^= {(a + n— 1.6)(a + n6) ... (a + w + r — 1 .6)}.
Then
v^ — t;„_j = {(a + n — 1 ,b)(a + nb) ...(a + w + r* — 1 .6)}
{(a + n2.6)(a + n1.6)...(a + n + r2.6)}
= {(a + n1.6)...(a + w + r2.6)}{(a+n+r1.6)
(a + n^.b)}
= (r+l)6{(a+nl .6)...(a + w+r2.6)}.
SUMMATION OF SERIES. 897
Hence v^ — v^.^ = (r + 1) 6 x u^.
Changing n into n — 1 we have in succession
Also Vj — v^ = (r + 1) 6 X Wj,
where t;^ is the term preceding v^ which is formed accord
ing to the same law,
that is t;^, = {(a — 6)a(a + &)...(a + r — 1 6)}, so that v^ is
obtained by putting n = in the expression for v^.
Hence by adjiition
Ex. 1. Sum the series 1.2 + 2.3 + 3.4+ +n(n+l).
Here u^=n{n+l), t;^= n(n+l)(n+2), 17^ = 0.1.2, r = 2, and
6=1.
Henoe iSf = ^ n (n + 1) (n + 2).
o
Or, by using the above method without quoting the result, which
is preferable in yerj simple cases, we have
n(n+l)=K{w(w+l)(n+2)(nl)n(n+l)},
3
1
3
(nl)n=i{(nl)n(n+l)(n2)(nl)n}.
1.2={1. 2.80. 1.2}.
Hence 5^=511 (n+l)(n+ 2).
o
Ex. 2. Sum the series 1.2. 3+2.3.4 + +n(n+l) (yt+2).
Ans. ^n(n+l)(n+2)(n+3).
398 SUMMATION OF SERIES.
Ex. 8. Bom the series
1.2.8.4+2.8.4.5 + +n(n+l)(n+2)(n+3).
Aru. gn(n+l)(n+2)(n+3)(n+4).
Ex. 4. Find ihe sum of n tenns of the series
8. 5.7+5.7.9+7.9.11 +
Here
i«,=(2n+l)(2n+3) (2/i+6), t;,=(2»+l) (2ii+3) (2ii+6) (2n+7),
t;<j=l. 3.5. 7, r=3, and 5=2.
Hence iSr^= ji^{(2n+l) (2n+3)(2n+5) (2n+7)l. 8. 5 . 7}.
Many series which are not of the requisite form can be
expressed as the algebraic sum of a number of series
which are all of the required form ; and the sum of the
given series can then be written down. The following are
examples.
Ex. 5. Find the sum of n terms of the series 1.8+2.4 + 3.5 +
Here 11^=: n (n + 2) = fi (n + 1) + n.
The sum of the series 1.2 + 2. 8 + ...+n(n+l) is
i{n(n+l)(n+2)0.1.2},
and the sum of the series 1+2+ ... +nis;r{n(n+l)0.1}.
Hence the required smnis>n(n+l)(n+2)+j^n(n+l).
Ex. 6. Find the sum of the series
2.3. 1 + 3.4.4+4.5.7+ + (n + l)(n+2) (3n2).
Here «^=(n+l) (n+2)(3n2)=3n(n+l) (n+2) 2 (n+1) (n+2).
.. fif^= {n(n+l) (n+2) (n+3)0. 1 .2.3}
{(n+l)(n + 2)(n+3)1.2.3}
=i(9n8)(»i + l)(n+2)(n+8) + 4.
SUMMATION OF SERIES. 399
319. To find the stmt of n terms of the series whose
general term is
l/{(a + n1.6)(a + n6)(a + n + 1.6)...(a + n + r2.6)}.
Consider the series which is formed according to the
same law but with one factor taken away from the
beginning of each term, and let v^ be the nth term of this
second series, so that v^= 1/ {(a + n6)...(a + nHr — 2 .6)}.
Then
1
^nVl =
{(a + w6) ... (a + n + r  2 . 6)}
1
{(a + 71  1 . 6) (a + w6) . . .(a + w + r  3 . 6)}
[(a + nl . h)
{(a + n 1 . 6). . .(a+ n+r— 2 . h)]
(a+n + r2.6)};
•'• «'nVi = (^l)*x Un
changing n into w — 1 we have in succession
Vi  ^'.a =  (^  1 ) ^ X ^Hi >
Vj — Vj = — (r — 1) 6 X w^.
Also Vj — Vq = — (r — 1) 6 X t^j,
where v^ is the term which precedes v^ and which is formed
according to the same law, that is
v^ = 1 / {a (a + 6).. .(a + r^2 . 6)}.
Hence, by addition,
t;,t;, = (rl)6xfif,;
.•.'S,=K0/(^i)6.
Ex.l. Siimtheserie8^3 + 3^+... + ^~jj^^^^.
400 SUMMATION OF SEBIES.
°*°^ ^*=r7i 12"^ =2""^T2
11 i
Ex. 2. Sum the series , ^ ^ ^ + ^ » ^ > + ... +
1.2.3.4 2.3.4.6 n(n+l)(n+2)(n + 3)
to n terms and to infinity.
XT 1 1
Here tt«= — ;^ =r; ^rr; rr , r«=
'»
n(n + l)(n+2)(n + 3)' * (n+l)(n+2)(« + 3) '
^0=1" 2^ » *'=^» and 6=1.
Hence ^» jj , i jf ^ g . 3 (n+1) (n+2)(n+3) *
^ « 1 1 1
Ex. 3. Sum the series ^ „ tt + „ ,, ..^ + .. . +
3.7.11 7.11. 15^'" ' (4»l)(4n + 3)(4»+7)'
Aw, ^»=8 jo "" (4n+3) (4n+7)f '
Many series which are not of the above form can be
expressed as the algebraic sum of a number of series
which are all of the required form; and the sum of the
series can then be written down. The following are
examples.
Ex.4. Sumtheseries =— 5+^r— 7 + rp + ...
Here
1 71+1 1 1
«n=
n(n + 2) n(n + l)(n + 2) (n+l)(n+2) ^»(n + l)(n + 2)'
The series whose general terms are . ttt ^ f^^
(n + l)(n + 2) n(n+l)(n + 2)
are of the required form. Hence the sum of the given series is
given by
« ^\_Ji\j.\lJi 1 \_3 2n + 3
* ^2 n + 2y "^2^1.2 (« + l)(n + 2)y~4 2(n + l)(n + 2)'
SUMMATION OF SERIES. 401
Ex. 5. Snm the series „ ^ .. + tr— ; — t+ ... +
Wn=
1.3.6^2.4.6^"*^n(n+2)(n+4)*
1 (n + l)(n+8)
w(n+2)(n+4) n(n+l)(n+2) (n+3) (n+4)
w(?i+4) + 3
n(n + l)(n+2)(n + 3)(n + 4)
1 3
(w + l)(n+2)(n+3) w(n + l)(n + 2)(n+3)(n+4)'
Hence
g^liJ: 1 1 3 ( 1 1 ]
* 2(2.3 (7i + 2)(n + 3)J ■'■4 11.2.3.4 (n+l)(?i + 2)(n + 3)(n+4)j '
320. The sum of series of the kind just considered
may be obtained by means of partial fractions.
The method will be seen from the following example.
To find the sum of the series r—^ + ttj + Tr?+ ••• +
1.3 2. 4 ^3.5^ ••^71 (n + 2)*
1 A B
Let —. rrr = — "i ', thcH, as in Chapter zxiii, we find
7i(n + 2) n n+2
that A= and B= .
^ 11
Hence 2w= —
* w n+2*
We have therefore the following series of equations :
oil nil oil
2^=13' 2^=2"4' 2^=3"6'
llo 11 ^oll
n — 55 n * TO — 1 TO+1 n to+2
Hence, by addition,
2S.44 ^
'*'"1^2 TO+1 TO + 2'
the other terms all cancelling.
„ 3 2TO+3
Hence 'Sf =   ^r; :rr, — .
* 4 2(to + 1)(to+2)
321. To find the swrn of the rth powers of the first
n whole nvmbers.
We will first consider the two simplest cases.
aA. 26
402 SUM OF SQUARES AND CUBES.
Case L To find the sum of l* + 2* + 3*+ ... + n*.
Here t*^ = w* = n (w + 1) — w.
Hence, by Art. 318,
/S» = n(n+l)(n + 2)n(n + l)
= n(w + l)(2n + l).
Case IL To find the sum of 1' + 2' + 3* + ... + w*.
Here u^ = n^ = n(n + 1) {n ] 2) Sn*  2n
= n (n + l)(w + 2) 3/i(w + 1) + w.
Hence, by Art. 318,
fif^ = l7i(n + l)(w + 2)(n + 3)n(n+l)(n + 2)
+ n(w+l)
= ^7i(n + l){(w + 2)(n + 3)4(7H2) + 2}
Since 1 + 2 + ... +71 = 2^(71 + 1),
the above result shews that
l» + 2»+...+7i» = (l + 2+... + ny,
so that the swa of the cvhes of the first n whole numbers is
equal to the square of the sum of the numbers.
The sum of the cubes of the first n integers can also be easily
found by means of the identity 4afi = {n (n + 1) }*  {(w  1) n)K
For we have in succession
4w»={n(»+l)p {(w 1) n}a,
4(nl)8={(nl)w}a{(n2)(nl)}3,
4.23 = (2.3)a(1.2)a,
and 4.1«=(1.2)»(0.1)a.
Hence, by addition, 4iS^=v? (n+1)'.
SUM OF POWERS OF INTEGERS. 403
Case III. To find the sum of l*" + 2*" + S** + . . . + n*".
The sum for any particular value of r can be found by
the same method as that adopted for the values 2 and 3.
For example, the sum of the fourth powers can be
written down as soon as w* is expressed in the form
w* = w (n + 1) (n + 2) (rH 3)  6n(n + 1)(7H 2)
+ 7fi(n + l)n.
By means of the Binomial Theorem a formula can be
found which gives the sum of the rth powers in terms of
the sum of powers lower than the rth ; and this formula
can be used for finding the sum of the 2nd, 3rd, 4th, &c.
powers in succession. The formula has however the great
disadvantage that in order to find by means of it the sum
of the rth powers, it is necessary to know the sums of all
the powers lower than the rth.
By the Binomial Theorem, we have in succession
(,tr> = (n~ir^H(r+l)(niyH ^''^^^^'' (nl)^
+ ... + 1,
3^^ = 2^^ + (r + l)2' + ^^^i^2'"^+... + l,
Hence, by addition, we have (n + 1)*^^ — (w + 1)
= (r + l) 8: + ^^^^8r+ ...+{r + l)8:,
where /S/ is written for the sum of n terms of the series
r+2'^+3*+...
26—2
404 PILES OF SHOT.
We can in a similar manner find a formula for summing the
rth powers of any series of quantities a, a+b, a+2b, ... in arith
metical progression. The result is
(a+nft)'+idHi.,ar*i==(r+l)6V + ^Y^^^'^**"^"^*"
where 5^«'=ar+(a+6)'+... + (a+n^6)^.
322. Piles of Shot. To find the number of spherical
bails in apyra/midal heap, when the ha^e is (I) an equilateral
triangle, (II) a square, and (III) a rectangle.
I. In a pile of this kind the balls which rest on the
ground form an equilateral triangle, and upon this first
layer a number of balls are placed forming another equi
lateral triangle having one ball fewer in each side than in
the side of the base ; and so on ; a single ball being at the
top.
If n be the number of balls in each side of the base,
the total number in the base will be
71 + (n1) + (712)+.. .+ 2+1,
that is \n (n + 1). The whole number of the balls in the
pile will therefore be
i {71 (71 + 1) + (711)71+.. .+ 1.2},
that is ^n {n + 1) (71 + 2).
II. In this case the balls in any layer form a square
with one ball fewer in each side than in the layer next
below. Hence if n be the number of balls in each side of
the lowest layer, n^ will be the number of balls in the base,
and therefore the whole number of the balls will be
7i» + (71  1)* + (71  2)* +. . .+ 1*, that is jTi (71 + 1) (271+ 1).
III. In this case the balls in any layer form a
rectangle with one ball fewer in each side than in the
layer next below. Hence if n and m be the number of balls
in the sides of the lowest layer, nm will be the number of
balls in the base and therefore the whole number of the
balls will be, n being greater than m,
nm + {nl) (m l) + ( n  2) (7^1 ^ 2)+ ... (n m + 1) 1
a=(7l — 77l + 77l)7W + (7l — m + TW.— l){m — 1) + . . .(tI — 771 + 1) 1
FIGUBATE NUMBERS. 405
= (wm){m + (ml)+...+ l} + m' + (ml)'+...+ l" '
= ^ (n — m) m (m + 1) + ^ (m + 1) (2m + 1)
= im(m + 1) (3/1 m + 1).
Ex. 1. How many balls are contained in 8 layers of an nnfinished
triangular pile, the number in one side of the base being 12 ?
If the pile were completed it would contain ^.12.13.14 balls;
and there are 7*4.5.6 missing from the complete pile; hence the
required number is ^ (12 . 13 . 14  4 . 5 . 6).
Ex. 2. How many balls are contained in 10 layers of an incomplete
pile of balls whose base is a rectangle with 20 and 25 balls in its
sides?
The number =2n (71+ 5) from n=ll to n=20.
Am, 3260.
323. Flgurate numbers. Series of numbers which
are such that the nth term of any series is the sum of the
first n terms of the preceding series, all the numbers of
the first series being unity, are called orders of figurate
numbers.
Thus the diflferent orders of figurate numbers are : —
First order, 1, 1, 1, 1, 1,
Second order, 1, 2, 3, 4, 5,
Third order, 1, 3, 6, 10, 15,
It follows from the definition that the nth term of the
second order of figurate numbers is n; the nth term of
the third order will therefore be (1 + 2 + 3 + . . . + n), that is
^n (n + 1); the nth term o{ the fourth order will therefore be
If/ . i\ . / i\ . . 1 o) xu X • n(n + l)(n + 2)
i{n(n + l) + (nl)n+...+ 1.2}, thatis— ^^ ^^ ^ ;
the nth term of the fifth order will therefore be
^{n(n + l)(n + 2) + (n^l)n(n + l)+...+ 1.2.3}, that
406 POLYGONAL NUMBEBS.
1
ig n(n + l)(n + 2) (n + 3); and so on, the nth term
of the rth order being
w(n + l)(w+2)...(n+r2)
r1
324. Polygonal numbers. Consider the arithmetical
progressions whose first two terms are respectively 1, 1 ;
1, 2; 1, 3; 1, 4; and so on. Then the series formed by
taking 1, 2, 3,..., n of the terms of these different arith
metical progressions, namely the series
1, 2, 3, , w,
1, 3, 6, , in(n + l),
1,4, 9, , <
1, 5, 12, , n+f7i(nl),
1, r, 3r— 3, ..., w + ^n(w — l)(r — 2), ...
are called series of linear, triangular, square, pentagonal,...
rgonal numbers.
The sum of n terms of a series of rgonal numbers
can be written down at once, for the sum of n terms of the
series whose general term is w + ^ (n — 1) (r — 2) is
^(n + l)H^(nl)7i(n + l)(r2)[Art.318].
EXAMPLES XXXIII.
Find the sum of n terms of each of the following series,
and find also the sum to infinity when the series is convergent.
1. 4.7.10 + 7.10.13 + 10. 13.16+...
Oi 1 1 1
^ 3.7.11 ■*'7.11.15"^11.15.19"^'"'
3. 1.3. 4 + 2. 4. 5 + 3. 5.6 + ...
4. 1.5 + 3. 7 + 5. 9 + 7. 11 + ...
EXAMPLES. 407
6. 1.2. 3 + 2. 3. 6 + 3. 4. 7+4. 6.9+...
6. 1.2» + 2.3« + 3.4« + 4.5«+...
7. 1.3» + 3.5« + 5.7»+7.9«+...
1 1 1 1
1.3.7"*'3.6.9"*'5.7.1l"^7.9.13^'**
1 1 1 1
• 1.3.4"^2.4.5"^3.5.6"*'4.6.7"*""*
10 ^ 6 6 7
1.2.3"^2.3.4"*'3.4.5"*"4.5.6^ '•
n ^ 2 3 4
1.3. 6"*" 3. 5. 7"*" 6. 7. 9"*" 7. 9. 11^"
12 ^ 4 5 6
1.2.4'*"2.3.5"^3.4.6'**4.5.7"^*"
IQ 1 1 1 1
^^' T"*'TT2'*'l + 2 + 3'*"l + 2 + 3 + 4"**
,^ V P + 2» l« + 2« + 3« l«+2*+3' + 4«
14. j^—^^ 3 + 4 +...
16. l.P + 2(l« + 2«) + 3(l«+2« + 3")+4(r + 2' + 3« + 4^+...
16. a' + (a + by + (a + 26)'+ ...
17. a* + (a + 6)' + (a + 26)«+...
18. r+3» + 6» + 7« + ...
19. l« + 5" + 9«+13' + ...
20. Shew that
l«__2' + 3«4'+...+(2w + iy = (n+l)(2?i+l).
21. Shew that 1" 2" + 3« 4»+ ...  (2n)" =  w (2*i + 1).
22. Shew that
l»2»+3»4»+...+(27i+l)»=4w» + 97i'+6w + l.
23. Find the sum of the series
1 .w+2(nl) + 3(w2) + ... +n. 1.
408 KXAMPLKS.
24. Find the sum of the series n . n+ (n  1) (n+ 1) +
(n  2) (n + 2) + . . . + 2 (2w  2) + 1 . (271  1).
25. Find the sum of n terms of the series
a6 + (al)(5l) + (a2)(62) + ...
26. Prove that, if /^/ = 1' + 2'' + ... w"; then wiU
(i) 6^/=6.Vx^/^/.
(ii) 8:^s:^2{s:)\
27. Find the sum of the following series to n terms :
3 1 4 1 5
,.., _3_1 ^J_ 5 1
..... 4 /2\ 5 /2V 6 /2\'
<"'^ 0(3)^273(3) ^ob)^
<'^> 0:1 (7) ^ 0:4 (7) +3:4:5(7) +•••
, , 9 /3\ 10 /3\* 11 /3V
<^) 073 (4) + 2:3:4 (4) ^ 30(4) ^ 
. ., 15 /6\ 16 /6V 17 /6V
<^*> 1:^(7) ■" 273:4(7) +34:5(7) +"•
28. Shew that the sum of all the products of the first n
natural numbers two together is ^ (w — 1) w (w + 1) (3w + 2).
29. Shew that the sum of all the products of the first n
natural numbers three together is j^ (?i  2) (w  1) w' (w + 1)*.
30. Shew that the sum of the products of every pair of the
squares of the first n whole numbers is
gl^n(n'l)(4n'l)(5«+6).
SUMMATION OF SEBIEa 409
325. To find the sum of n terms of the series
a (a + x). . ,(a + n — 1 x)
"7"  — "■ ~y' • • •
6 (6 + a?). . .(6 + n — 1 x)
In the above series there is an additional factor both
in the numerator and in the denominator for every succes
sive term, and the successive factors of the numerator and
denominator form two arithmetical progressions with the
same common difference.
Consider the series formed according to the same law
but with an additional factor in the numerator, and let v^
be the general term of this second series, so that
^n =
a(a\ x),..(a + n — 1 a?) (a + nx)
6 (6 + a?). . .(6 + w — 1 x)
Then
Vn'y«,=
_ a(a + x), ,,(a ^ n — 1 x) (a { nx)
6 (6 + a?). . .(6 + n — 1 x)
a(a + fl?)...(g + n — 1 x)
b{b + x)...(b + n'2x)
= T7I r~7I — ==^](a + nx)'{b+nlx)[ ;
b(b+x).,.(b + n'lx)[ J
.*. ^'w— 'I'l.i = ^n X (a + a?  6).
So also v^^ — v,^ = u^^ X (a + a? — 6)
v,— Vj = t^j X (a + a? — 6).
Also v^=a — r—^ = {a + x)u^
= M, X (a + a? — 6) + bu^.
410 SUMMATION OF SERIES.
Hence i8f,x(a+« — 6) = r^ — a;
The sum of n terms of the series
a a(a—x) a(a — a?)(a — 2a;)
T
b 6(6 + a?) • b(b + x)(b + 2a;)
•••
in which the successive factors of the numerator and
denominator form two arithmetical progressions whose
common differences are equal in magnitude but of opposite
sign, can be found by changing the sign of a in the
previous result : the sum can, however, be obtained inde
pendently by the same method. Thus
a_ 1 Ta a(a — a?)"l
b^a + bxll^ 6 J
_ a{a'x) _ ^ 1 fa (a — a?) a{a—x){a — 2x) '\
b{b + x) a + bx [ b "*" b(b+x) J
bib+x)...(b + ^r^x)
/ I \«i r a(ga?)...(an— 1 x)
^ lb{b + x)...{b + ^'^2x)
b{b + x)
Hence
a{a — x).,.{a — nx) " 
...(b+n — lx)}
• a + 6a;L ^ ^ 6(6 + a;)...(6 + 7rn:a;) J'
Ex, 1. To find the sum of n terms of the series +—^ + ^' ^' ^h
8 8.6 3.6.9
SUMMATION OF SERIES. 411
We have
21 /2, 5 2\ 2.5_l/ 2.5.8 2.5\
3~2V 3 "ly' S.6~2V 3.6 3 J
2.5.8...(3wl) _ 1 f 2.5.8...(3n+2) _ 2. 5. 8... (3n 1) 1
3.6.9...3n ""5 I 3.6.9...3n 3.6.9...(3n3)J '
Hence 5,=^ ____ _^ .
[This partioalar series is a binomial series, the successiye terms being
the coefficients of x, x\ &c., in the expansion of (laj)~*. Hence
[Art. 287] 1 + ^H=B^^"^ ^' ^^^ ^^^ (^+^) coefficients in the expansion of
(1  a:)"* = coefficient of aJ* in (1  a?)"* x (1  x)~^, that is in (1  a?)"^].
„ ^ „. ,^, , . .^, .. 2 2,6 2.6.10
Ex. 2. Find the sum of « terms of the series  + ^— ^ + 5^r^rr + • • •
o t5.7 o.7*ll
3 I3.7...(4n1) j
Ex. 3. Find the smn of n terms of the series
 m m{ml) m (m  1) (m  2)
T^~TT2 17273 **"•••
Ans. (,l)Hi(^"l)(^"g)(^^+l).
1 . 2,..(n — 1)
326. The sum of n + 1 terms of the series
«o + ^1^ + «i^ + ... + a»^*,
where a, is any integral expression of the rth degree in w>
can be found in the following manner.
Hence Sf. x (1  a?)'^' = ao + {a,  (r + 1) aj a? +. . .
f (r h 1) r )
+ jap  (r + 1) ttp^j + 1 2 ^^«~'"[ ^'^ "^ •'•
+ (~ ir ^ a ^^
n
412 SUMMATION OF SEBIEa
Now a, is by supposition an integral expression of the
rth degree in p ; hence
% = A^p'+A^^,f' + A^p^ + ... + A,,
where A^y J.^^,..., A^ do not contain p,
Ako, by Art 305, the sum of the series
/(r + l)()~l)* + ^^^±^(p2)*...to(r+2) terms,
is zero for all integral values of k less than r + 1. Hence
(^7*+ 1)7*
ap(r + l)a,.i + .. ^ a^^ ... to (r + 2) terms
is zero for all values of p.
All the terms of the product S^ x (1 — a?)*^* will there
fore vanish except those near the beginning, or the end,
for which the series %— {r +1) a^^+.., is not continued
for (r + 2) terms, that is all the terms of the product will
vanish except the first r + 1 terms and the last r + 1 terms.
Hence
8^ X (1 a?)'^*= a, + {a, (r + 1) a^} a?+...
r / . V (r + 1) r .
a;"**
whence the value of 8^ is found.
Ex. 1. Find the sam of the series
l+2ar+3a;2+4x'+ +(n +!)«•*.
fif^i=l + 2x + 3a;» + 4x»+ +(w+l)x*,
(lx)a=l2a;+a;2;
[aU the other terms Yanishing on acconnt of the identify
*2(*l) + (*2)=0]
«=l(n+2)ar«+H(n+l)a;'»+a;
. Q _1_ (n + 2) a;*+^  (n+ 1) g*"*^
SERIES WHOSE LAW IS NOT GIVEN. 413
Ex. 2. Find the sum of n + 1 terms of the series
V+2^x + S^x^'{' +{n+lfx\
5r^i=l« + 28x + 88a;a+ + (n+l)»a;»
(lx)*=l4a? + 6a;a4aH»+a:*;
.'. 5„4.iX(lj;)*=l + (28.4)x + (3»4.2»+6.13)a;3
+ (484. 33+6.284. 13) a;8
+ {4(n + l)8 + 6n84(nl)8+(n2)8}a;"+i
+ {6 (to + 1)8  4n8 + (n  1)'} a;»+2
+ {4(n + l)8 + n8}a:»^8
+ (n + l)8j;»*+^.
[The other terms all vanishing, since
Jk84(*l)8+6(A:2)84(A;3)8+(*4)8 = identically.]
Hence iS'^i=[l + 4a?+a;2(n8+6n«+12n+8)a;«+i
+ (3n8 + 16na + 21n + 5) a;*«
 (3n8 + 12n2 + 12n + 4) x*^
+ (n + l)8a;*W]/(la;)4.
When X is numerically less than 1, the series is convergent, and the
snm of the series oontinned to infinity is (1 + 4a; + a;") / (1  x)\
327. Series whose law is not given. We have
hitherto considered series in which the general term was
riven, or in which the law of the series was obvious on
inspection. We proceed to consider cases in which the
law of the series is not given. With reference to series
in which the law is not given, but only a certain number of
the terms of the series, it is of importance to remark that
in no case can the actual law of trie series he really deter
mined : all that can be done is to find the simplest law the
few terms which are given will obey.
There are for instance an indefinite number of series
whose first few terms are given by x + a? +a^ + ,.., the
simplest of all the series being the geometrical progression
whose nth term is x* : another series which has the given
terms is that formed by the expansion of —
1x
10
414 METHOD OF DIFFERENCES.
which agrees with the geometrical progression except at
every 10th term.
Note. In what follows it must be understood that
by the law of a series is meant the simplest law which
satisfies the given conditions.
Method of Differences.
328. If in any arithmetical series
each term be taken from the succeeding term, a new
series is formed, namely the series
which is called the^r^^ order of differences.
If the new series be operated upon in the same way,
the series obtained is called the second order of differ
ences. And so forth.
ThuB, for the seriee 2, 7, 15, 26, 40, ... ,
the first order of difierences is 6, 8, 11, 14, ...,
and the second order of differences is 3, 3, 3, ...
329. When the law of a series is not given, it can often
be found by forming the series of successive orders of
differences ; if the law of one of these orders of differences
can be seen by inspection, the law of the preceding order
of differences can often be found, and then the law of the
next preceding order of differences, and so on until the
law of the series itself is obtained. The method will be
seen from the following examples.
Ex. 1. Find the nth term of the series
1 + 6 + 23 + 58+117+206+
The first order of differences is 6 + 17 + 36 + 59 + 89 +
„ second „ „ „ 12+18 + 24+80+
„ third „ „ „ 6+6 + 6 +
METHOD OF DIFFERENCES. 4*15
The second order of differences is clearly an arithmetical progres
sion whose nth term is 6 (n+ 1).
Hence, if v^ he the nth term of the first order of differences, we
have in succession
Also Vi= 6 . 1  1. Hence, hy addition,
v^=6(l2+ +n)l = 3n(n + l)l.
Then again, we have in succession tt^  u^_i = v^_i = 3 (n  1) n  1 ;
*'niWna=3(w2)(nl)l; ...; t*2Mi=3. 1.2 1. Alsot^ = l.
Hence w^=3 {{nl)n+ + 1 .2}n+2 = (nl)n(n+l)n + 2.
Ex. 2. Find the nth term and the sum of n terms of the series
6 + 9 + 14+23+40 +
The first order of differences is 3 + 6 + 9 + 17 +
„ second,, „ ,,2 + 4 + 8+
Hence the second order of differences is a geometrical progression,
the (n  l)th term being 2"~^. Hence, if v,^ be the nth term of the
first order of differences, we have in succession
^»«nl=2*•~^ v^i'V^2=^'''^ Vat;i=2i.
Also Vi=3. Hence, by addition, v,»=(2 + 2a+ + 2»*i) + 3=2*+l.
Then again, we have in succession w„w,j_i=v,j^i=2*~i+l,
^ni"^Ha=2*~* + l , tig 1*1=2^ + 1. AlsoUi=6.
Hence w^=(2»i+... + 2)+n+6=2»+n+3.
The sum of n terms of the series can now be written down : for
the sum of n terms of the series whose general term is 2"+n+3 is
(2+2a+...+2»*) + {n+(nl) + ... + l} + 3n=2«+i_2+in(n+l)+8n.
Note. By the method adopted in the preceding
examples the nth term of a series can always be found
provided the terms of (me of its orders of differences are all
the sa/me, or are in geometrical progression,
330. It is of importance to notice that when the
nth term of a series is an integral expression of the rth
degree in n, all the terms of the rth order of diflferences
will be the same.
416 RECUBRINO SERIES.
For, if u^ = A;rf+ A^^7r^\ .,. + A^, where A^y A^_^,...
do not contain n, the nth. term of the first order of differ
ences will be
[A^(n+ir+A^,{n + Vr'+...]^[AX + A^,7r'\...},
which only contains n to the (r — l)th degree.
Similarly the nth term of the second order of differ
ences will be of the (r — 2)th degree in n ; and so on, the
nth term of the rth order of differences being of the (r — r)th
degree in n, so that the nth term of the rth order of
differences will not contain n, and therefore all the terms
of that order of differences will be the same.
When therefore it is found that all the terms of the
rih. order of differences are the same, we may at once
assume that u^ = A^rH' + A^^riT^ + . . . + A^, and find the
values of J.^, A^_^y . . ., J.,, by comparing the actual terms of
the series with the values obtained by putting n = 1, n = 2,
&c. in the assumed value of u . This method will not
however give the value of u^ m a convenient form for
finding the sum of the series ; for, if r be greater than 3,
the sum of n terms of the series whose general term is
A^nT {• A^^vT^ + ... cannot be found [see Mt. 321] without
a troublesome transformation which will in fact reduce u»
to the form in which it is obtained by the method of
the preceding Article. A much better method would be
to assume that u^ = A^ (n\ + A^^ Wri + •••> and then to
find 4^, 4^j,..., A^ as above.
Recurrinq Series.
331. Definitions. When r + 1 successive terms of
the series a^ + ajc + ajif + + a„a?** +. . . are connected by
a relation of the form a^ a?* + px (a^.j a?""*) + qx^ (a ^ a?*"*)
+ ... = 0, the series is called a recurring series of tne rth
order y and l+px\q(if\ ... is called its scale of relation.
The relation does not hold good unless there are r terms
before the nth, so that the relation only holds good after
the first r terms of the series.
RECURRINQ SERIES. 417
For example, the series l + 2ic + 4r' + 8a?'+ is a
recurring series of the first order, the scale of relation
being 1 — 2a?. Again, it will be found that the series
1 + 3a? + 6aj* + 7a?' + 9a?* + is a recurring series of the
second order, the scale of relation being 1 — 2a? + a?*.
332. To find the sum of n terms of a given recurring
series.
Let the series be a^ h a^x + + ajxf* +. . ., and let the
scale of relation be 1 + pa? + qa^. [This assumes that the
recurring series is of the second order, but the method is
perfectly general]. Then
since all the other terms vanish in virtue of the relation
(ij,a? +px (aj_i a?*"^) + qa? (aj^^ a?*'*) = 0, which is by sup
position true for all values of k greater than 1.
Hence
* l+px\qa?
If the given series be a convergent series, the nth term
will be indefinitely small when n is increased without
limit; and the sum of the series continued to infinity
will in this case be given by
g ^ ap + idt+pdo)^
* 1+px + qa^
The expression °^ ^ ^ — ^ ^ is therefore such that if it
1 +px 4 jar
can be expanded in a convergent series proceeding accord
ing to ascending powers of x, the coefficient of af in its
expansion will be the same as in the recurring series.
s. A. 27
418 BECUBEINQ SERIES.
On this account the expression "., . ' — . \ is
1 +px + qscr
called the generating fwnction of the series.
333. A recurring series of the ith order is determined
when the first 2r terms are given.
For let the series be
©o + ^i^ + ^s^ + +a„a? +
Then, the series being a recumng series of the rth
order, if we assume that the unknown scale of relation is
1 +j[>^x\p^a?+.,,+p^af, we have by definition the follow
ing equations
«r+8 + PiCtrirl +2> A + . . . + PA =0,
=0,
We have therefore r equations which are suflScient to
determine the r unknown quantities p,,^,, ...,2)^ in the
scale of relation ; and when the scale of relation is deter
mined the series can be continued term by term, for a^^.^
is given by the equation aj^i + p^ag^ + . . . H ^^a^ = ; and
when ttg^^j is found, a,^,^ can be found in a similar manner ;
and so on.
The series is similarly determined when cmy 2r con
secutive terms are given.
334. From Art. 305 we know that iip<r + l,
to r + 2 terms = 0,
for all values of k.
BECURBING S£RI£S. 419
This shews that the series
is a recurring series whose scale of relation is (1 — ic)'^\
It also shews that the series
is a recurring series whose scale of relation is (1 — xf^^
whenever a^ is a rational and integral expression of the
rth degree in n.
335. In order to find the sum of any number of terms
of a recurring series by the method of Art. 332, it is neces
sary to know the general term of the series; we must
therefore shew how to obtain the general term of a
recurring series when the first few terms are given.
By Art. 333 the scale of relation of a recurring series
of the rth order can be found when the 2r first terms are
given ; and, having found the scale of relation, the genera
ting function is at once given by the formula of Art. 332.
Now, provided the scale of relation can be expressed
in factors of the first degree, the generating function can
be expressed as a series of partial iractions of the form
A A
or of the form 7= rv, and the coeflScient of any
1—OUV (100?)*' "^
power of a? in the expansion of the generating function
can be at once written down by the binomial theorem;
and thus the general term of the series is found.
When the value of x is such that the given recurring
series is not convergent, the generating function will not
be equal to the given series continued to infinity nor can
it be expanded in a series of ascending powers of x ; but,
taking as an example the generating function in Art. 332,
the expression ^^ ^ ^ — ^ %^ can always be expanded in
ascending powers of y, if y be taken sufficiently small, and
27—2
420 RECURRINQ SEBIEa
the coefficients of jf and y* in this expansion will clearly
be a^ and a^ respectively and all succeeding terms will
obey the law a^ •VpOj^x + <l<hrJi = 0, and hence all the coeffi
cients of the expansion will be the same as the corre
sponding coefficients in the given series. We may there
fore in all cases, whether the series is convergent or not,
find the general term of a recurring series by writing
down the expansion of its generating function in ascending
powers of x on the supposition that x is sufficiently smalL
Ex. 1. Find the nth term of the reonrring series 8 +4a;+ 6^+ 10x*+ ...
In an example of this kind, in which the order of the recurring
series is not given, it mnst always be nnderstood that what is wanted
is the reonrring series of the lowest possible order whose first few
terms agree with the given series. In the present example there is
a sufficient nmnber of terms given to determine a recurring series of
the second order, but an indefinite number of recurring series of the
tldrdf or of any higher order than the second, could be found whose
first four terms were the same as those of the given series. [See
Art. 327.]
Assuming then that the scale of relation is l+j'^+ga:', we have
the equations 6+4P + ^=0> <^d 10 + 6p +4g=0, whence >s 8 and
g = 2. Hence the scale of relation is 1  3a; + 2ac^.
The generating function is therefore
3 + (49)a? _ 35a? 2 1
l3a; + 2a?»"~l3a;+2a;2~la;"**l2a;
=2 {!+«+ ... +a;*i} + {1 + 2a; + ... + 2»ia;*»i + ...}.
^ence the general term of the series is (2 + 2*"^) x^^K
The sum of n terms can now be found by the method of Art. 332;
the sum can however be written down at once, for the sum of n
terms of the series 2(l+a;+a:^+...) is 2 (1  a;*») / (1  a?) and the sum
of n terms of the series 1 + 2a?+4ic^+ ... is (1  2*a;*»)/(l  2a;).
We may remark that the given series is convergent provided a; <^.
Ex. 2. Find the nth term and the sum of n terms of the series
1+3 + 7+13+21+31 + ....
Consider the series l+3a;+7a;'+13aH*+21a;*+8LB"+...
Then, assuming that the series is a recurring series, and also that
a sufficient number of terms are given to determine the recurring
series completely, it follows that the series is of the third order.
Let then the scale of relation be l+p^ + gaj^+ra;'; we then have
the following equations to find Ptq^rt
RECURRING SERIES. 421
13+7p+3gr+r=0,
21 + 13p + 7g+3r=0,
and 31 + 21p + 13g + 7r=0,
whence p=3, ^=3 and r=l,
so that the scale of relation is 1  3ir + Sx^  a^.
The generating function is now found to he
l+g» _ 2 2_ 1_
{1xf^iXxf {lx)^'^ix'
Hence the general term of the series
l + 3a;+7ar»+... isa;*i{n(n+l)2n+l} = (7i2n+l)a;»i.
Thus the general term of the given series is n^  n+ 1.
Having found the general term of the series the sum of the first n
terms can he written down, for the sum of n terms of the series
whose nth term is n (n 1) + 1 is ^ (n 1) n (n+ 1) +n.
Ex. 8. Find the nth term of the series 2 + 2+8 + 20+
Considered as a recurring series of the lowest possihle order, the
generating function of 2+2a;+8a;3+20a;i^+... will he found to be
22a;
l2a;2a;a
Now the factors of 1  2a;  2x^ are irrational, and therefore the
nth term of the series, considered as a recurring series of the second
order, will he a complicated expression containing radicals.
On the other hand, by the method of Art. 329, we should be led
to conclude that the nth term of the series was (3n^  9n + 8) jb*^^,
which by Art. 334 is a recurring series of the third order.
As we have already remarked, the actual law of a series cannot be
determined from any finite number of its terms, and the above is a
case in which it would be difficult to decide as to what is the
timplest law that the few terms given obey, for the recurring series
of the lowest order which has the given terms for its first four
terms is not the recurring series which gives the simplest expression
for the nth term.
CONVERGENCT AND DIVERGENCY.
336. We shall now investigate certain theorems in
convergency which were not considered in Chapter XXI.
422 CONVERGENCY.
337. OoiiTergency of infinite products. A product
composed of an infinite number of factors cannot be con
vergent unless the factors tend to unity as their limit ; for
otherwise the addition of a factor would always make a
finite change in the continued product, and there could be
no definite quantity to which the product approached
without limit as the number of factors was indefinitely
increased.
It is therefore only necessary to consider infinite pro
ducts of the form
n(l+u,) = (l+wJ(l + w^(l + W3)...(l+wJ,..,
where t^ becomes indefinitely smsdl as ?i is indefinitely
increased; and the convergency or divergency of such
products is detennined by the following theorem.
Theorem. The infinite 'prod'wfA n (1 + w^), in which
all the factors are greater than unity, is convergent or
divergent according as the infinite series Xu^ is convergent
or divergent.
Since e* > 1 + a?, for all positive values of x, it follows
that
(l+t^,)(l + M^(l+wJ...<e»».^.^...<^+««+*»+
Hence, if Xu, be convergent, n (1 + u^) will also be
convergent.
Again, (1 + wj(l + w^ > 1 + Wj + w„
and so on, so that
n(i+u;)>i + %u^.
Hence, if Xu^ be divergent, H (1 + u^) will also be
divergent
Ex.1. Toaiewi^i l^'['^]\^^/^^J^^^^^
6(6 + l)(6 + 2)...(6 + nl)
n is indefinitely increased, according as a is greater or less than b.
CONVERGENCY. 423
For, if a > &, the ezpresBion may be written in the form
.('*^)(;i0 (ss^i)
which is greater than
But ^ + r — 7 + X — o + ... is a divergent series [Art. 274] : the given ex
pression is therefore infinite when n is infinite, a being greater than b.
If 6>a: then as before, —7 — 777 tzt—^ is infinite; and
' a(a + l) (a+2)
therefore^i^±Jli^±^L:::i: must be zero.
Ex. 2. Determine whether the series
41 a{a+x) a(a\'x) (a + 2a;)
h'^ b{b + x)^ b{b+x)(b + 2x)'^
is convergent or divergent.
From Art. 325, we have
_ a i (a+x){a+2x)„,{a + nx) A
''~'a+xb\i,^fi + x)(b + 2x)...(b + n~'~i,x)' ]'
XT 1 11 I {a + x)(a + 2x)...{a+nx) , . « .. ,.
Now by Ex. 1, ^^ ^ — ■■ —  —  is infimte or zero according
b(b + x)...(b + n'lx)
as a + « ^ 6*.
Hence the given series is convergent, and its sum is then ,
~' CL '" X
if 6 > a+o;. Also the series is divergent if & < a+x.
Also if 6=a+a?, the series becomes z + i ^t — tt which is
b b+x b+2x
known to be divergent [Art. 274].
338. The Binomial Series. We have already proved
that the binomial series, namely
is convergent or divergent, for all values of m, according
as a? is numerically less or greater than unity.
If a? = 1, the series becomes
1 . ^ , m(ml) m(ml)(m2)
424 CONVERGENCT.
Now we know that the terms of this series are
aUemately positive and negative after the rth term, where
r is the first positive integer greater than m + 1. More
over the ratio u^Ju^ is numerically less or greater than
unity according as m + 1 is positive or negativa The
series will therefore, fiom theorem V. Chapter XXL be
convergent when m + 1 is positive provided thei nth term
decreases without limit as n is increased without limit.
^^ . 1 1.2 n
Now ± — =
u^ (— m) (1 — m). . .(ti — 1 — 77i)
u^ m\ l—mj\ 2 — m/ V n — l — mj
Now, if m + 1 be positive and less than r, the product of
the factors from the rth onwards is greater than
(1+^)_J_ + _1 +.. I.
^ ^ {r^m r + 1 — m J
and the product of the preceding factors is finite.
Hence, when n is increased without limit, 1/m^ is in
finitely great, and therefore u^ indefinitely small, provided
1 + m be positive.
Thus the binomial series is convergent i/" a? = 1, pro
vided m > — 1.
If a? = — 1, the series becomes
n _ . m(ml) m(ml)(m2) ,
^ "^ 1.2 1.2.3 ■^••"
The sum of ri terms of the above series is easily found
to be [see Art. 287 or Art. 325]
(1 ~m)(2 — m)(3m)...(yil — m)
1.2.3...(nl)
The sum of n terms of the series is therefore [Ex.
1, Art. 337], zero or infinite, when n is infinite, according
as m is positive or negative.
Thus the binomial series is convergent when a? = — 1,
provided m is positive.
CONVERGENCY. 425
339. Cauchy'8 Theorem. If the series u^\u^ + u^
+ ... + ^^ 4 ... have ail its terms positive, and if each term
be less than the preceding, then the series will he convergent
or divergent according as the series v^\aUa^a^Ua^ + ,,.
+ a"i^«» + .,.is convergent or divergent, a being any positive
integer.
For, since each term is less than the preceding, we
have the following series of relations
Wj + «^8 + ... + w« < aMi < (a — 1) 1^1 + Wi,
Hence, by addition, 5<(a — l)S + Wi (I),
where 8 and S stand for the sum of the first and second
series respectively.
Again, we have since a is ^r 2,
a (^1 + t*g + Wj +... + t*J > au^
Hence aS>%'U^.,, (II).
From I and II it follows that if 8 is finite so also is S,
and that if 8 is infinite so also is X.
Ex. To shew that the series —7; rr is convergent if A; be greater
n (log 71)* "
than nnity, and divergent if ft be equal or less than unity.
By Gauchy's theorem the series will be convergent or divergent
a«
according as the series whose general term is ^. ^ is convergent
or divergent.
426 CONVBRGENCT.
Now S °* =S^ ^ =: ^ Z^ '
a* (log a*)* n* (log a)* (log a)* n* '
it therefore follows from Art. 274 that thagiTen series is conyergent
if iSe> 1 and diyergent if iSe:^ 1.
340. We shall conclude with the two following tests
of convergency which are sometimes of use, referring the
student to Boole's Finite Differences and Bertrand's Differ
ential Calculus for further information on the subject.
341. Theorenu A series is convergent when, from
and after any particular term, the ratio of each term to the
preceding is less than the corresponding ratio in a known
convergent series whose terms are all positive.
For let the series, beginning at the term in question,
be
U — u^ + u^^ u^+ .., ^u^^ ...,
and the known convergent series, beginning at the same
term, be
yyj + ^,+ ^3+ ... + V„ + ....
U V
Then, since ^* < ^^ for all values of r, we have
u^ Vr
> V, + V, ^ + i^.   + i^i  —  + . . .
>l'(u, + u, + u^ + u, + ,..)>^U.
Hence as F is convergent, U must also be convergent.
The given series is therefore convergent, for the sum
of the finite number of terms preceding the first term of
U must be finite.
We can prove similarly that if, from and after any
particular term, u^^ : u^ > v^, : v^, and all the terms of Xu^
have the same sign ; then 2w^ will be divergent if Xv^ be
divergent.
CONVBRGENCY. 427
342. Theorem. A series, all of whose terms are posi
tive, is convergent or divergent according as the limit of
n
( 1 2±i j is greater or less than unity.
For let the limit of w (1 !^+M be a.
n
Consider the series S s = Sv. : then
nr *
/ v^A _ ((n + 1)^ — n^\ _ pn^ + lower powers of n
\ v» / "" I {n + iy j "" n^ + lower powers of n
Hence the limit of nil —^], when n is infinitely
great, is p.
First suppose a > 1, and let ^ be chosen between
a and 1.
Then since the limit of nil ^M is greater than
the limit of n ( 1 — 5+m there must be some finite value
of n from and after which the former is constantly greater
than the latter.
But when ^(l  ^)>^(l ^)»
we have ^* > ^^ .
^n ^n
Hence, by the previous theorem, 2t^„ will be conver
gent if Sv. be convergent; but %v^ is convergent since
/S>1.
Similarly, if a be < 1, and ^ be taken between a and
1, we can prove that 2t^„ is divergent if 2v„ is divergent,
and the latter series is known to be divergent when /8 < 1.
If the limit of n ( 1 ^M be unity the test fails.
428 EXAMPLES.
■n. ^ T XI. • « a(a+l) a(a+l\((n+2) ,
Ex. 1. IB the Bene, + ^J^*+ Jj^Zi__J^+... convergent
or divergent ?
Here **^ = r «, the limit of which is x. Hence, either from
u^ o+n
the beginning or after a finite nomber of terms,
^^ ^ 1 according as a; ^ 1.
Hence the series is divergent if « > 1, and convergent if jb < 1.
If xsl, the limit of ^^ is nnity. Bat
"{''^)"('S)'
the limit of which is &  a.
Thus, if x=:l, the series is convergent when ba>l and
divergent when 5  a < 1. When &= a + 1, the series becomes
a a a
b'^bTi'^b+2'^ '
which is divergent.
[These are the results arrived at in Ex. 2, Art. 337.]
EXAMPLES XXXIV.
1. Find the sum of each of the following series to n
terms, and when possible to infinity : —
r\  ?L^ 2.5.8
.... 3 3.5 3.5.7
^^"^ 8"*"8.10"^8.10.12"^*'"
. 11 11.13 11.13.15
^^""^ 14 "*■ 14. 16 "^14. 16.18 "^•••*
2. Find, by the method of differences, the nth term and
the sum of n terms of the following series : —
(i) 2 + 2 + 8 + 20 + 38+....
(ii) 7 + 14 + 19 + 22 + 23 + 22 + ...,
(iii) 1 + 4 + 11 + 26 + 57 + 120+....
EXAMPLES. 429
(iv) 1 + + 1 + 8 + 29 + 80 + 193 + ....
(v) 1 + 5 + 15 + 35 + 70+126 + ....
(vi) 1+2 + 29 + 130 + 377 + 866 + 1717 + ....
3. Find the generating function of each of the following
series on the supposition that it is a determinate recurring
series : —
(i) 2 + 4a;+14a? + 52a?"+....
(ii) l + 3a?+lla?' + 43aj*+....
(iii) 1 + 6a? + 40aj*+ 288a;" + ... .
(iv) 1 + a? + 2a* + 7a;" + 14a?* + 35a;* + ....
(v) 1" + 2^x + 3V + 4V + 5V + 6V + ....
4. Find the nth term, and the sum of n terms of the
following recurring series : —
(i) 2 + 6 + 14 + 30+....
(ii) 25 + 2989+....
(iii) 1 + 2 + 7 + 20+....
5. Find the nth term of the series 1, 3, 4, 7j <!^c. ; where,
after the second, each term is formed by adding the two
preceding terms.
6. Determine a, 6, e, c? so that the coefficient of a^ in
XI. • ^a + hx+cx^ + da? , , v.
the expemsion of .^ ^ .^ may be (n + 1)'.
7. Shew that the series 1'+ 2''a? + 3V + 4V+ ... is the
also that a, = 0; and that tfr.,= a _i.
8. Find the sum to infinity of the recurring series
2 + 5a? + 93?* + 15a? + 25a?* + 43a?* + ...
supposed convergent, it being given that the scale of relation
is of the form 1 + jw? + qpi? + tt?. Shew that the (n + l)th term
of the geries is (2" + 2n + l)a?".
430 EXAMPLES.
9. Find the sum to infinity of the series
1 + 4aj+ llaj*+ 26a' + 57aj*+ 120aj"+ ...,
X being less than ^.
10. Find the sum of n terms of the series
 X x(x + a) x(x + a)(x + b)
a ao aoo
IL Shew that
la ah
+ 1 n tT + 7 w rr7 V + ...
05+ a {x + a){x + b) (x + a) (a? + 5) (x + c)
abc..,k 1 ahc.,,hl
{x + a){x'^h)...{x + k){x + l) X x(pc{a){x + b) ...{x + l)'
12. Shew that
1 1+w (l+w)(l+2w)
jE)+w (j9 + w) {p + 2») ( JO + w) (p + 2w) (p + 3n)
+ ... to infinity = — =
provided that /? > 1 and /? + w > 0.
13. Shew that, if m be greater than 1,
. 1 1.2 1.2.3
1+ 7 +
r/* + l (m+l)(w» + 2) (m+l)(m + 2)(w + 3)
+ ... to infinity = =■ .
14. Shew that
_1 n1 (nl)(n2) 1
m+1 (m+l)(m + 2) (m + l)(m + 2)(w> + 3) *"""w+w'
if 9i» + n be positive, or if n be a positive int^er.
15. Shew that, if ti be any positive integer,
n n{n—l) w (n  1) (n  2)
^T+T " (n + l)(n+2) ■** (n+l)(w+2)(n + 3)'** ••*
w(nl)(n2)...2.1 1
"^ (w+l)(n + 2)...2n "2'
EXAMPLES. 431
16. Shew that, if m be a positive integer,
2n+l m{ml){2n+l){2n + 3)
^~^2n + 2'*" 1.2 (2n + 2)(2w + 4) •'•
1.3.5...(2wl)
(2n + 2) (2n + 4) ... (2n + 2m) '
17. Shew that, if m, n and w — w + 1 are positive integers;
then
 m w(n — 1) m(m — 1)
m — n+l 1.2 (w — w+ l)(m — w + 2)
n(n — \)(n—2) m(m— l)(m— 2)
1.2.3 {mn+l) (w — n+ 2) (mw + 3)
+ to^ni lUormn (m+ 1) (m+ 2)...(m + m)
^ ' (mw+ l)(w» — »+ 2) ... (wiw + m)
18. Shew that, if m + 1 > 0, then
1 1 lm(m— 1) 1 w>(m — l)(m— 2)
2"'3'^'*"i"~T:2 6 1.2.3 ^•"
1
■"(m + l)(m+2)*
19. Shew that, if P^ be the sum of the products r together
of the first n even numbers, and Q^ be the sum of the products
r together of the first n odd numbers; then will
1 + Pj+P,+ +P. = 1.3.5...(2n+1),
and l + ^j + ^^4 + C. = 2.4.6...2n.
20. Prove that
{a + (a+l) + (a + 2) + ...+(a + n)}{a' + (a+l) + (a + 2)+...
... +(a + n)}«a* + (a+l)'+... +(a + n)*.
21. Shew that the series
. lg" (1  g") (1 ~ g"') (1  oT) (1  g" ' ) (1  a "«)
1g "*" (lg)(la») (1  a) (1  a') (1  a») "*"••'
is zero when n is an odd integer, and is equal to (1 — a) (1  a")
... (1 — a"~^) when n is an even integer [Gauss].
432 EXAMPLES.
22. Find the sum of the series
n w— 1 n—2
1.2.3 2.3.4 3.4.5 n{n + l)(n + 2y
OQ a X . ^ x 2aj» 3> 4a?*
23. Sum to infinity =— «  zr—: + s— e  ....
1 . o 2,4 3.5
24. Sum, when convergent^ the series
172 "*■ 273 ■*■••• ■*"^^+T) "^ ••• '
25. Sum to infinity the series
1 . 2 . 3 + 3 . 4 . 5a? + 5 . 6 . 7aj" + 7 . 8 . 9a;^ + ...,
X being less than unity.
26. Shew that, if n is a positive integer
i3...grL^)^(3"^^)f«).....=2(ir.
27. Shew that, if a^, a^, a^,... be all positive, and if
a^ + a^ + a^+ ... be divergent, then
^ ■"(«.+ !)(».+ 1)+ (a, + !)(«,+ !)(«.+ 1)+
is convergent and equal to unity.
28. Shew that the series
1 2"* 3"* w"
is convergent if aj> 1, and is divergent if a? I^ 1.
29. Shew that, if the series u.+u+u+.,.+Un+... be
divei'gent, the series
*+ — *— + ... + = + •..
will also be divergent.
t:xAMPL£:s. 433
30. For what values of x has the infinite product
(1 + a) (1 + 005) (1 + ax*) (1 + ax^) ... a finite value]
31. Prove that, if v^ is always finite and greater than
unity but approaches unity without limit as n increases
indefinitely, the two infinite products VjV^v^v^...^ v^v*v*v^,..
are either both finite or both infinite.
32. Test the convergency of the following series : —
,.. 1 2' 3" nr
W 2^ + 3* + 4* "*■•••"*■ (/i + 1)+* "^ ••••
r'\ 1 _L _L _L
^"^ 1 "*■ ^2* "^ 4/3* "*"'*•■*" '^n"'"'' "^ ■*• *
r'\ A ^'^ 2.4.6
^"^^ 3.4'*'3.5.6"*"3.5.7.8"^"*
2. 4. 6. ..271
3.5.7...(2n+l)(2» + 2)
. 1 1.3 1.3.5
^^""^ 2.3'^2.4.5'^2.4.6.7'*''"
1.3.5.. .(2wl)
■'■2.4.6...2w(2»+l)'^""
ap a(a+ 1)^08+1)
'^ ' l.y"^ 1.2.y(7+l) "*■•••
a(a+l)(a+2)^(^ +1) (/8 + 2)
1.2.3.y(y+l)(y + 2) ■^•■•'
S.A. 28
CHAPTER XXVL
INEQUALITIES.
343. We have already proved [Art. 232] the theorem
that the arithmetic mean of any two positive quantities is
greater than their geometric mean. We now proceed to
consider other theorems of this nature, which are called
Inequalities,
Note. Throughout the present chapter every letter
is supposed to denote a real positive quantity.
344. The following elementary principles of inequal
ities can be easily demonstrated :
I. If a > 6 ; then a\x>b\ x, and a — x>b^x,
II. If a > 6 ; then — a < — 6.
III. If an> b ; then ma > mb, and — ma < — mb.
IV. Ifa>6, a'>6', a''>r,&c.;
then a + a' fa" +... >6 + 6' + 6"+...,
and aa!a!' .., > bb'V\...
V. If a>6; then a'">6^ and a'^< 6"~.
Ex. 1. Prove that a^+l^'>a%\al^.
We have to prove that
a«a«6a6«+68>0, or that (a*  6«) (a  6) > 0,
which most be true smce both factors are positive or both negaiiye
according as a is greater or less than h.
INEQUALITIES. 435
Ex. 2. Prove that a*^ + a~*» > a* + a~*, if m > n.
"We have to prove that (a^  a*) (1  a~^~*) > 0, which must be
the case since both factors are positive or both negative according as
a is greater or less than 1.
Ex. 3. Prove that (P + m' + n^) (Z'« + m'^ + w'^) > {W + mm' + nn')^.
It is easily seen that
(^2 + m» 4 n2) (Z'a + m'^ + n'«)  (Zr + mm' + rin')^
= (mn'  m'n)2 + {nV  n'lf + (Im!  Vmf.
Now the last expression can never be negative, and can only be
zero when mn'  mfn^ nV  n'l and Imf  Vm are all separately zero, the
conditions for which are rr = 7 = 7 .
Hence (P+m'+n2)(Z'2+m'2+n'2) > (M'+mm'+nn')^, except when
l\V ■='m.\w! —n\n\ in which case the inequality becomes an equity.
345. Theorem I, The 'product of two positive quanti
ties^ whose sv/m is given, is greatest when the twofax^tors are
equal to one another.
For let 2a be the given sum, and let a+ x and a — x
be the two factors. Then the product of the two quanti
ties is a" — a?y which is clearly greatest when x is zero, in
which case each factor is half the given sum.
The above theorem is really the same as that of Art. 232 ; for
from Art. 232 we have (^) (^) > «&•
346. Theorem II. The product of any number of
positive quamiities, whose sum is given, is greatest when the
quantities are all equal.
For, suppose that any two of the factors, a and b, are
unequal.
Then, keeping all the other factors unchanged, take
J(a + b) and ^{a + b) instead of a and b : we thus, without
altering the sum of all the factors, increase their continued
product since J(a +b)x ^(a + 6) > ab, except when a = 6.
Hence, so long as any two of the factors are unequal,
the continued product can be increased without altering
the sum ; and therefore all the factors must be equal to
one another when their continued product has its greatest
possible value.
28—2
436 INEQUALITIES.
Thus^ unless the n quantities a^b, c, ... are all equals
,, /a + 6 + c + d+ ...V
abed... <{ 1 ,
and therefore
n
> ^{ahcd ...).
By extending the meaning of the terms arithmetic
meam, and geometric mean, the last result may be enunci
ated as follows : —
Theorem III. The arithmetic mean of any number of
positive quantities is greater than their geometric msan,
Ex. 1. Shew that a» + 6* + c* > babe.
We have ^ > i/(a*^h». c») > dbe.
Ex.2. Shew that ^ + ^ + ^+ + ^>n.
a^ Ot a^ di
We have i(^ 4^+ 4M > C^ f" •••. V^l
Ex. 3. Find the greatest valae of (a x) {b y) {ex + dy), where a, 5, c
are known positive quantities and a  x, by are also positive.
The expression is greatest when (aeex)(bddy){cx+dy) is
greatest, and this is the case, since the sum of tne factors is now con
stant, when accx=bddy=ex + dy. Whence the greatest value is
found to be (ac + 6d)'/27cd.
Ex. 4. Find when x^y^z^ has its greatest value, for different values of
Xf y and z subject to the condition that x+y+zia constant.
Let F^x'^y^z^; then
a,
a
XXX y y y z z z
aaa P P P 777
The sum of the factors in the last product is constant, since there
X v z
are a factors each , 8 factors each ^, and y factors each , and
therefore the sum of all the factors ia x+y+z.
INEQUALITIES. 437
Hence, from Theorem II, (j f^j fj has its greatest value
when aU the factors are eqnaL that is when  = ^ =  .
a /3 7
It is clear that P is greatest when Pja^^y* is greatest, since
a, /3, 7 are constant; hence P is greatest when x I a=ylfi=zly.
In the above it was assumed that a, /3, 7 were integers ; if this
be not the case, let n be the least common multiple of the denomina
tors of a, pf 7. Then x'y^z^ will have its greatest value when
g^y'^z^ has its greatest value, which by the above, since 71a, np
X V z
and ny are aU integers, will be when — = ^ = — > that is when
' ® na n/3 n7
0/37*
Thus, whether a, ^, 7 are integral or not, x''tfz* is greatest for
values of x, y and « such that x + y + 2 is constant, when x/a = ylp=.z\y,
347. Theorem IV. The swm of any number of
positive quantitieSy whose product is given, is least when
the quantities are all equal.
First suppose that there are two quantities denoted by
a and 6.
Then, if a and 6 are unequal, {s/a * V^)* > 0, and there
fore a + 6 > Jab + ^/a6. Hence the sum of any two
unequal quantities a, b is greater than the sum of the two
equal quantities Jab, Job which have the same product.
Next suppose that there are more than two quantities.
Let a, b, any two of the quantities, be unequal. Then,
keeping all the others unchanged, take Jab and Jab
instead of a and b : we thus, without altering the product
of all the quantities, diminish their sum since Jab + Jab
< a+ 6. Hence, so long as any two of the quantities are
unequal, their sum can be diminished without altering
their product; and therefore all the quantities must be
equal to one another when their sum has its least possible
value.
438 INEQUALITIES.
348. Theorem V. //* m and r be positive, and m > r;
then
* '^ ^ will he greater than
n n
mr
n
We have to prove that
n (a,"* + a,** +...)> (a/ + a/+ ...) (a^""^ + a,""^ + ...)»
or that
(nl)(a,«' + a," + ...)>S(aX"'' + a,"'^0,
or that
2 (a  + a   a/a,"'  ^i^XO > 0,
every letter being taken with each of the (n — 1) other
letters.
Now
a  + a,"  «^  a ^'a; = (a/  <) (a ""^  a."^,
which is positive since a^*" — a^ and a^"*^ — a,"*"^ are 6o^
positive or both negative according as a^ is greater or less
than a,.
Hence S (a * + a *  a^"^  ai"*"^0 > 0,
which proves the proposition.
By repeated application of the above we have
ta^ tal 2< 2V
71 71 71 n
where a, /8, 7, ... are positive quantities such that
a + )8 + 7 + =771.
Ex. 1. Shew that 3 (a»+6«+c») > (a + 6+c) (a'+i^ic*).
Ex. 2. Shew that a»+6»+c»>a6c(a«+6»+c2).
rrom Theorem y, e!±^> ^±^ . (i±±^)*,
> ^'+^+<^ . a5c, from Theorem m.
o
INEQUALITIES. 439
349. Theorem VI.* To prove that, if a, b, c, .., and
a> A 7, ... be all positive, then
First, let a,b,c, ... be integers. Take a things each a,
b things each 13, and so on. Then, by Theorem III,
(g + a+ ... to a terms) + {13 + ^+ ... to b terms) H ...
a + 6 + ...
^««H^J^«^r, },
that is ' ^"!"t^,^"' >"^V{g"iQ''}'
If a,b,c, ... be not integral, let m be the least common
multiple of the denominators of a, 6, c, ... ; then ma, mb,
mc, ... are all integers, and we have
macL\mb^+ ... ^ ^+„u^ , i^ma^i, .
Hence { ^ + f+ l'^" >«.^^ [A].
1 1
Cor. I. Put a = , iS = r , ..., and let there be n of the
a
letters a, 6, ... ; then
0+B+.. I
>
ta + 6 + ...J ^a«6\..'
Cor. II. Substitute in (A) a*" for a, 6'' for 6, . . . ; also
substitute a"*"^ for a, 6"*"^ for /8, ..., where m>r.
* See a paper by Mr L. J. Rogers in the Messenger of Mathematics^
Vol. XVII.
440 INEQUALITIES.
Again, substitute a'', Vy ... for a, 6, ... respectively, and
a*"*", 6'"^, ... for a, )8 respectively, where t<r.
Then
r/i*" 4 fe*" 4 ) a'+6'+...
M^^rfe:} <K^"r [c].
Hence, as m — r and r — t are both positive, we have
from PB] and [C]
Hence, provided m>r>t,
{a"'+6"'+...px{a'"+6'+...}*~x{a*+y+...}"'>l....T)].
The following are particular cases of [D].
Put t = 0; then, since a" + 6**+ ... = n, we have provided
m>r
{^^^}'>{^^tH m
Again, put t=0,m=l; then since m>r>ty r must
be a proper fraction.
Hence, if r be a proper fraction,
f a + 6 + >.. r ^ a'' + y+... jp,
I w j n *■ ■**
Again, put < = 0, r = 1 ; then m > 1.
Hence, if m> 1 t«;e Aaw
a* + 6'" + ... fa + 6+...l"' ^^^
Now put m = 1, r = 0, then t is negative. Hence, pro
vided t be negative,
{a + b+ ...y* X rT^ X (a* + b* h ...)>0;
a*+b'+... /a + 6 + ...V
n \ n
) m
INBQUALrriES. 441
From [F], [G] and [H] we see that
n < L ^ J
according as a; is not or is a proper fraction.
350. We shall conclude this chapter by solving the
following examples. [See also Art. 133.]
Ex. 1, Shew that, if a=ai+aa+... + ajj,
9 S 8 vS^
1 4.... +  > — _. unless a,=aQ=...=a.
Unless Oi=aj=...=a^, we have
1 / » % * \ * / **
and^ — ^^—^ — ^ ^^ ^>M(»ai)(««a)(<«»)}
By multiplication, since (sOi) + (aOj) + ... + («aJ = iw»,
we have
n1 / « s » \ ^
5 ( + +... + )>1.
Ex. 2. Shew that, ifa + & + c + (2=3«; then will
abed > 81 («  a) {s  6) («  c) (a  d).
For 8^{(a6)(»c)(»d)}<{(«6) + (ac)+(ad)}<a.
Soalso 3^{(»c)(«d)(aa)}<6, 34/{(»d)(»a)(»6)} <c,
and 3^{(aa)(a~6) (8c)}<d.
Hence 81 («a)(»6) (»c) («d) <a5c<i
^ J > ae'yifz^, unless x=y=^g.
First suppose that a;, y and z are integral; then by Theorem m.
{X + X + ... to a; terms) + (y+y + ... to y terms) + (g+g4.. . tog terms)
a? + y+«
... (!±^')*^>^y.^.
\a;+y+«/ ^
If X, y, « be not integral let m be the least common multiple
of their denominators ; then mx, my and ms are integral, and we
have by the first case
\ mx+my+nus / \ / v ;f/ \ / ^
442 EXAMPLES.
/a? 411* 4 «'\**t^i'+*>
that is ( ^ 1 X m"»<»Hr+^ > («*t/i'z')"» x m"»<*+«^+^;
\ x+y+z J
The Theorem can in a similar manner be proved to be true for any
number of quantities.
EXAMPLES XXXV.
Prove the following inequalities, all the letters being
supposed to represent positive quantities : —
2. (a/ + a/ + a,*+ )(5/ + 5/ + 6/+ )
4. a'^ + h'^a'b + ah'.
\a cj \x y zj ^
6. (a + 6 + c) (a" + 6" + c*) ^ 9a6c.
7. a*cc? + h^ad + c^oi + d^hc ^ 4a5ccf.
8. (6c + ca + ahy ^ 3a6c (a + 6 + c).
9. a* + 6* + c* <t a6c (a + 6 + c).
10. a* + 6* 4 c* + fl?' ^ a5a? (a + 6 + c + <f).
_ a — 05 a'— aj* .
IL < ^ — j, , II a; < a.
a + a; a' + ar
1 1 1 1 1 1
a 6 c ^yjoc Jca ^ ao
13. SheT^ that, if a5^* + a:,'+ +aj/ = a,
then Tia > (ajj + 052 + + oj^)* > a.
EXABfPLES. 443
14. Prove that, if ajj, x^, x^ , x^ be each greater than
a, and be such that (aj^ — a) (aj^ — a) {x^ — a) = b% the least
value of x^x^x^,,.x^ will be (a + 5)*, a and h being positive.
15. Shew that {a^^ a^^
,o 2 2 2.9
16.
17.
b+c c+a a+h a + b i c'
3 3 3 3 . 16
b + c + d c + d + a d + a + b a + b\c^a + b+c+d'
18. Shew that if a > 6 > c; then
\a — c/ \6 — c/
19. If a;* = y* + «•, then will x*^y* + z* according as w ^ 2.
20. Shew that (ubcd)''*'*^^^ lies between the greatest and
1111
least of a', 6*, (f, d' .
. 21. Shew that l+aj + a:'+ +aj" ^(2n+l)aj".
22. If n be a positive integer, and a > 1 ; then
n—sz — =— >
23. (m + l)(m + 2)(m + 3) (m+ 2nl)<(77Hw)^"\
24. abc <^ (6 + ca)(c + a6)(a + 6c).
25. abed <i(i (b + c + d2a) {c + d + a — 2b) (d + a + b 2c)
(a + 6 + c2fl?).
where (»— l)« = aj + aj+ +a,.
27. If a, 5, c be unequal positive quantities and such that
the sum of any two is greater than the third, then
1119
+ 7 + 
b+c—a c+a—b a+b—c a+b+c'
444 EXAMPLES.
28. Shew that^ unless a = 6 = c,
(6^c)«(6 + ca) + (ca)'((j + a6) + (a6)'(a + 6c):>0.
29. Shew that, if a, 6, c be unequal positive quantities,
then
a«(a6)(ac) + 6*(6c)(6a) + c'(ca)((j6):>0.
30. Shew that ^"'' + ^af "' + raf "* >^ + j + r, unless a;=l,
or p = qssr.
oi ai^ XI. X 1.3.5...(2nl) / 1
^^ «^^"*^* 2.4.6..2n' V2;rrT'
on ai. XI. X 3.7. 11. ..(4711) /
32. Shew that ^ ^ ,^ — h ^ < . /
5.9.13...(4»+1) V ■
4w + 3"
33. Find the greatest value of x^'y^z'^, for different values
of aj, y, and z subject to the condition that ax^ + by^ {csi* ^d,
34. Prove that, if » > 2, ([w)* > n\
35. Shew that, if n be positive,
(l+a;)(l+aj*)>2"'V.
36. In a geometrical progression of an odd number of
terms, the arithmetic mean of the odd terms is greater than
the arithmetic mean of the even terms.
37. Prove that, if an arithmetical and a geometrical pro
gression have the same first term, the same last term, and
the same number of terms; then the sum of the series in a. p.
will be greater than the sum of the series in G. p.
38. Shew that, if P^ denote the arithmetic mean of all
those quantities each of which is the geometric mean of r out
of n given positive quantities; then P,, P^, ..., jP, are in
descending order of magnitude.
39. Shew that, if « = a+6 + c+ ...,
\nl) \nl) Ul/ '"^W'
n being the number of the unequal positive quantities a, 6, c, ... .
40. Shew that, if w be any positive integer.
CHAPTER XXVII.
Continued Fractions.
851. Any expression of the form a±b
c±d
e ± &c.
is called a continued fraction.
Continued fractions are generally written for con
venience in the form
^h d f
o± e± g ±
352. The fraction obtained by stopping at any stage is
called a convergent of the continued fraction. Thus a and
a ±  , that is j and — =^ , are respectively the first and
h /J
second converffents of the continued fraction a A —  —  ...
^ ""c ± e ±
The rth convergent of any continued fraction will be
denoted by ^ .
h fJ
The fractions a, , , &c. will be called the first,
c e
second, third, &c. elements of the continued fraction.
446 CONTINUED FRACTIONS.
b d
353. In a continued frdction ofthefonfii a+
where a, 6, c, <fcd. ar« aK positive, the convergents are
alternately less and greater thom ike fraction itself.
For the first convergent is too small because the part
— ... is omitted; the second convergent, a + , is too
C "T* " c
great because the denominator is really greater than c;
then again, the third is too small, because c +  is greater
e
than c^ —  ...; and so on.
e +
354. In order to find any convergent to a continued
fraction, the most natural method is to begin at the
bottom, as in Arithmetic : thus
c^ a, a^^ g, ^ afifi^ + a^a^
fci + ^2 + 63 , aj>^ bfi^b^ + bfi^ f a^ftj *
If only one convergent has to be found, this method
answers the purpose ; but there would be a great waste of
labour in so finding a succession of convergents, for in
finding any one convergent no use could be made of the
previous results: the successive convergents to a continued
fraction are, however, connected by a simple law which we
proceed to prove.
355. To prove the law of formation of the successive
convergents to the continued fraction
a+^i ^ ?L8
^ + 6. + 6» +
The first three convergents will be found to be
a ab^ + a, ^^^ abfi^ + aa, + a A
1 ' b, bA + a.
CONTINUED FRACTIONS. 447
Now the third convergent can be written in the form
(ab^ + g J b^ + (a) a,
6, . ftj + 1 . a,
from which it appears that its numerator is the sum of the
numerators of the two preceding convergents multiplied
respectively by the denominator and nu/merator of the last
element which is taken into account / and a similar law
holds for the denominator.
We will now shew by induction that all the convergents
after the second are formed according to the above law
provided there is no cancelling at amy sta^e.
For, assume that the law holds up to the nth con
vergent, for which the last element is a^^Jb^^, and let
Priir denote the rth convergent ; then by supposition
Then the (7i + l)th convergent will be obtained by
changing t^* into , **"* r^ , that is into r — ^^^ — .
,11—1 »— 1 T /» 1 T 1
Hence in (i) we must put a^.^ b^ for a^_j and b^_^b^ + a^
for 6^.^; we then have
= K (KxPnl + ^nlPn^ + ««1>.1
= KPn+^nPnl [from i.].
Similarly gr„^, = b^q^ + a^q^,,.
Thus the law will hold good for the (w + l)th con
vergent if it holds good for the nth convergent. But we
know that the law holds good for the third convergent ;
it must therefore hold good for all subsequent ones.
1 1
Cor. I. In the fraction a, H — — ... ,
' (^2 + % +
Pn = ^nPni+Pn^ ^nd q^ = a^q^., f q„.,.
448 CONTINtJED FRACTIONS.
Cor. II. In the fraction r^ ^ ^ .♦. ,
Pn^KPni^nPn^ and q,^Kq^_^a^q^_^.
Ex. By means of the law connecting snccessive convergents to a
continued fraction, find the fifth convergent of each of the following
fractions :
1 1 1,1 .... 11111
^^f ^ + 1 + 2 + 5 + 4* ^"^ 4 + 1 + 4 + 1 + 4 +
•••
*••
/•••x 12346 ,..^2222
i^) 2 + 3 + 4 + 6 + 6 <^^> 3 + ^^^^^^^^
12345 /xlS?!!
^^^ 1 + 1+1+1 + 1* M^^> 4 + 3 + 2 + l + 2 + •••
, .., 2 2 2 2 2 , .... 11111
M 33333 (^)i. 4141
..•
356. The convergents to continued fractions of the
111
form a + r  j ..., where a,bfC,dy.,. are all positive
i c +a +
integers, have certain properties on account of which
such fractions have special utility: these properties we
proceed to consider. We first however shew that any
rational fraction can be reduced to a continued fraction
of this type with a finite number of elements.
For let — be the given fraction ; then, if m be greater
than n, divide m by w and let a be the quotient and p
the remainder, so that — = a 4 ~ . Now divide n by ©
and let 6 be the quotient and q the remainder; then
■^ = = — . Now divide p by g and let o be the
» « 6 + i
P P
quotient and r the remainder ; then " =  = — ^— . By
P P c+~
9 q
CONTINUED FRACTIONS. 449
proceeding in this way, we find — in the required form,
, m p 1 11
namely — =a+ = a + — — = a + ^ .  . ... •
•'w n 1 , Q 6+C +
P
Since the numbers jp, q, r, ... become necessarily
smaller at every stage, it is obvious that one of them
will sooner or later become unity, unless there is an
exact division at some earlier stage, so that the process
must terminate after a finite number of divisions.
It should be noticed that the process above described
is exactly the same as that for finding the G.C.M. of m
and n, the numbers a, b, c, ... being the successive quo
tients. On this account the numbers a, b, c &c. in the
1 1
continued fraction a + r  •• are often called the
6 + c +
first, second, third, &c. partial quotients.
It is easy to see that the continued fractions, found as
above, for —  and v , where k is any integer, will be the
same.
491
Ex. Convert 77^71 ^^^ 3*14159 into continued fractionSi and find in
1224
each case the fourth convergent. Ans, j=fj , yjo •
357. Properties of Convergents. Let the continued
fraction be a, H — ~ . ...» and let — denote the nth
convergent
L From Art. 355 we have
Pn Pn^X ^ <^n Pn^t + Pn^S Pni ^ Pn^ ^nl ~ Pnl ?»»« .
R. A. 29
450 ooNmiincD FRAcnoHS.
So also in soooession
PA PA =  (pa pad
But prf, p.g, = (a,a, + 1)  a.a, = L
Hence p.?^, p^,?. = ( 1)" (i)
Hence also &^' = tll)! (U).
11
Cob. In the continued fraction — — (•••.which is
less than unity, we have
IL Every common measure of p^ and q^ must also be
a measure of Pn^ni'' Pni^m ^^^ ^» bom I., a measure of
± 1. Hence p^ and q^ can have no common measure.
Thus all convergents are in their lowest terms.
III. IfF=a, + L 1 ... L ...; then^wiUbe
1 1
obtained from the nth convergent by putting —
in the place of — .
Hence ^ J'"^^, V>^^",l^±^ .
where X is written instead of — ... , so that \ is some
positive qua/rvtity less than wnity.
CONTINUED FKACTIONS. 451
Hence ^& = ?!l+2^._^.= M^^^sSitZEi?^)
_ (1)
,«l
3»(?i. + ^m.i)*
Also J^ ^*i = ^« "^ ^P«i ^"» = ( P**^*^ " ^*» ^«>)
Now \ is less than 1, and q^ is greater than q^^ ; hence
jP ^^ is less than F''^ .
\nrl
''■ Thus ariy convergent is nearer to the continued fraction
than the immediately preceding convergent, and therefore
' nearer than any preceding convergent,
^' IV. If any fraction,  suppose, be nearer to a continued
y
sc
fraction than the nth convergent, then  must from III. be
if
also nearer than the (n— l)th convergent; and, as the
continued fraction itself lies between the nth and the
(n — l)th convergents [Art. 353], it follows that  must
also lie between these conveigenta
Hence . ^' ~ must be <^» 2.
9.1 y 2«i 9.
i.e. i>..y~g,.^ <_i_.
9... y Inl^l
:. y must be > 9»(i'»iy  9t«')
Hence, as all the quantities are integral, y must be
greater than g, .
29—2
452 CONTINUED FRACTIONS.
Thus every fraction which is nearer to a continued
fraxytion than any particular convergent mast have a greater
denominator than that convergent
V. We have seen in III. that
where \ is a positive quantity less than unity.
Hence ^^> ^ .\^ . ;
also J^~25=t<_J_.
Thus a/ny convergent to a continued fra^ction differs
Jrom the fraction itself by a quantity which lies between
1 1
^j" ^^ T / f . TV , where d, and d^ are respectively the
demyrm/nxitors of me convergent in question and the next
succeeding convergent,
Ex. 1. Shew that, if p^jqf. be the rth oonyergent to the continaed
fraction ai + — — —.then will
p^i * a*i+ +aa + ai'
For we have l>i,=aa2>«i+Pna,
Hence ^ = a^+^«=?=a.+L = a^+ 1
P»l P%1 Pnl _ . I'nS
.11 1 Pi
« J ^ 111
EXAMPLES. 453
It can be proved in a similar manner that
n 111
Ex. 2. To shew that — = =5  . 5 ... to » quotients, where n is
w+i 4 — J — J —
a poeitiye integer.
We have
n 1 n1 1 2_ 1
'f ......f anQ ^^
n n— 1 2
Hence — = ^ ;r ;r to n qaotients.
n+l = 2 2  2  ^
Ex. 3. Shew that, if PriQr ^ ^^0 rth convergent of   ^
then will i)^i=ag^.
&li? ^* ^« Hence© aff
EXAMPLES XXXVI.
1. Shew that, ii ^.^,^he three successive conyersents
?. ^, ?a . **
to any continued fraction with unit numerators, then will
2. Shew that, if
— be the nth converirent of r* 7^ r^
then will p^q^^^ Pnx^n = ( 1)'"* «i«» ««•
3. Two graduated rulers have their zero points coincident,
and the 100th graduation of one coincides exactly with the
63rd of the other: shew that the 27th and the 17th more
nearly coincide than any other two graduations.
454 EXAMPLES.
4. Shew tbat, if a^, a,) , a^ be in barmomcal progression;
thenwiU— ^=Tr ^ ... ^ ^.
»., 22 2 a,
5. Shew that
111 ^ , 111
na.
6. Shew that, if P=  ^  ... ^
a +6 +c + '" + A;+ 1'
6 + c +<;? + '" +ife'
thenwiU P(a + ^+ !) = « + ©.
7. Find the value of
n n1 n2 2 11
w + »l +»— 2 + '" +2 + 1+2*
8. Shew that, whether n be even or odd, ^ t t t
'1—4 — 1—4 —
to w quotients = =• .
9. Prove that the ascending continued fraction
— — 5. ...IS equal to i +—^ + — =— + ...
»1 »8 «8 ^1 ^1^8 ^l^A
10. If p^ be the numerator of the wth convergent to the
fraction ^ ^ f^ .... shew that a linear relation connects
6, + 6, + 63 +
every successive four of the series p', p^^ P8*»***> ^^'^ ^^^
what the relation is.
11. If vJq^ be the rth convergent of  r  t ••
shew that ;?,^^, =^,, + 6g,,, and that q^^^^ = ap^^ + (aft + l)^^,^.
EXAMPLES. 455
12. If pjq^ be the rth oonvergent of the continued fraction
5 + F + J + Lf + J+ ••'*''*'' *^*^"=^'<^' ^>«
13. If pjq, be the rth convei^ent oi ^ J^ ^t ^\ ^ ....
shew that p„ q^_^  qt,p„.^ = a'b:
14. Shew that, if — be the 9»th convergent to the continued
fraction
15. Shew that, if = a + t  ...t t; then will
q 6+C+ k ■\ I'
11 11 1^1 1 1 l^JL^
I +A: + **' +c +6 +a I +Ar + '** +c +6 ~ pq'
p
16. Shew that, if ^ be converted into a continued fraction,
p
the first quotient being a, and the convergent preceding 2^ being
P . Q
 ; then, if — be converted into a continued fraction, the last
q' ^ q ^
convergent will be (P  aQ)/(p  aq).
17. Shew that, if and ^ be any two consecutive conver
gents of a continued fraction as, then will ^ ^ jb* according as
q<q"
358. To find the nth convergent of a continued
fracbUm,
We have in Art. 355 found a law connecting three suc
cessive convergents to a continued fraction, so that the
456 GENERAL OOKYEBGENT.
convergents can always be determined in succession. In
some cases an expression can be found for any convergent
which does not involve the preceding convergents : the
method of procedure wlQ be seen from the following
examples.
Ex. 1. To find the nth oonYergent of the continued fraction
1 1.3 3.5 6.7
...•
3+ 4 + 4 + 4 +
Here the nth element is ^ ^ — ^—^ , and therefore
4
J'n=^«ri + (2n  3) (2n l)i)»_3.
The above relation may be written
l'«{2n + l)i>^_i= (2n3) {Pn^i{2nl)p^^^}.
Changing n into n  1 we have in succession
Pni  (2n  l)>n2=  (2n  5) {p^^  (2n  3)i)^_,}.
P87P2=3{i»25i>i}.
But, by inspection, p=l, P2 = ^ J •'• l^j  5pi =  1.
Hence i),(2n+l)p^_j = (l)*i (2n3)(2»6)...3.1.
Then again
1.3...(2n + l)'"l.3...(2rtl)~(2»+l)(2nl)'
Pa Pi _(l)^
1.3.6 1.3"~3.5 '
'^^^ ^3=173
^^^ 1.3.6...(2n + l)""lT3'"3.6"** '*"(2n+l)(2nl)'
Since the denominators of convergents are formed according to
the same law as the numerators, we have from the above
g^(2n+l)^^_i=(~l)»3.6...(2n3){ga6gi} = 0,
since 9i=3 and q2=lB.
GENERAL CONVERGENT. 457
Henod
(2n + l)(2n~l)...3.1~{2»l)...3.1 ""5.3.1 3.1 '
.. g^=1.3...{2nl)(2n+l).
Hence pjqnt ^^^ ^^ convergent required, is
1 1 ^ ^ (1)"^
1.3 3.5^ ^(2nl)(2n+l)'
Ex. 2. To find the nth convergent of the continued fraction
12 3 4
l+2+3+i+
The necessary transformations are given in Ex. 5, Art. 251.
1 1 (1)"'
It wiU be found thati>»= J2 "" ]3 "*" "^ n+I~ *
andthat g,=l^ + . + ^_^.
If n be supposed infinitely great
!! 1
Pn^ ^
q^ 1  e"i e1'
359. Periodic continued fractions. When the
elements of a continued fraction continually recur in the
same order, the fiaction is said to be a periodic continued
frdction ; and a periodic continued fraction is said to be
simple or mixed according as the recurrence begins at the
beginning or not.
nru 111111
Thus a + T   r   . ...
: 1 A 1111
is a simple, and  r r r —
^ ' a + 6 + 6 + 6 +
is a mixed periodic continued fraction.
458 GSNSRAL OONYBBGENX.
860. 2V> ^n(2 ^ nt& (»mo07^en^ q/" a j)er»oc2ia continued
fradion with one recurring demerit,
h h h
Let the firaction be a +  .... Then, for all
C+C+C+
convergents after the second, we have p^ = (^^^^ + ^P»i
where 6 and c are constants, that is, are the same for all
valnes of n.
Now, if Uj +u^ + u^+...+t^^a^+...be the recurring
series formed by the expansion of ^ 5—=, the suc
1— ca? — oar
cessive coefficients after the second are connected by the
law tt, = cu^i + 6t*«Hr Hence, if A and B are so chosen
that tij=Pj and w,=l>j, then will w^=^^ for all values of
n. The necessary values of A and B are respectively p^
and p, — cp^, that is a and b.
Hence Uie numerator of the nth convergent to the
continued fraction a + ..... is the coefficient of a?*"* in
c +C +
, . « a + 6a?
the expansion of = r3 .
Similarly the denominator of the nth convergent is
the coefficient of «*"* in the expansion of ^' ^^* ^i~ ,
that is of
1— ca; — &i?'*
Ex. 1. Find the nth convergent of the continued fraction
3 3 3
The numerator of the nth convergent is the coefficient of a^"^ \
in the expansion of j^^, that is of 2^^j  gi^
Hence l>n=^{3*+(l)*}.
Also g,= coefficient of a?*~^ in the expansion of
__i _L . L
l2*&r»~4(l8a;) 4(l+a:)*
GENERAL CONYERQENT. 459
Hence g^=i{3« (!)»}.
Thus the nth convergent is 2 5= — 7 — ^ .
o* — ( — ly*
Ex. 2. Find the nth convergent of the continued fraction
a e a c
We have Pan=^Pami+cP9n2*
Pa*! = ^»i«+ ^ln8»
l'lM9 = ^Pam9 + ^9ii4 •
Hence, eliminating Pj^i ^^^ Pans* ^^ ^^^
P9^{a{e{hd)p^^+atp^^^=0.
Since the last result is symmetrical in a and c, and also in 5
and d, it follows that
Pti^i{ahc{bd)p^_^+acp^.f^=0.
. Hence the relation
Pn''{(Bt+c + hd)p^'^ <W!Ph4= ^
holds good for all values of n.
Hence Pf^ will be the coefficient of x*~^ in the expansion of
A+Bx + Cafi+Dx^
l(a+c + M)a:'+acx*'
provided the values of J , B, C, D are so chosen that the result holds
good for the first four convergents. It wiU thus be found that p^ is
the coefficient of «*'^ in the expansion of
a+adxac:i^
i(a+<j+6d)af*+ac«*'
It will similarly be found that 9^ ^ ^^ coefficient of x*~^ in the
expansion ^f
&+(6d+c)xacg*
i" (a + c + 6d) X* + OCX* *
361. Oonvergency of continued fractions. When
a continued fraction has an infinite number of elements it
is of importance to determine whether it is convergent or
not. When an expression can be found for the nth con
460 CONVEEGENCY OF CONTINUED FRACTIONS.
vergent, the rules already investigated can be employed;
the nth convergent cannot however be often found.
In the continued fraction r* . r* 5^ .... it is easy to
Oj + 6, + 6, + ^
shew, as in Art. 357, that
£2  Pari = (_ i)»i ^A'^i»
and hence that
Now, if all the letters are supposed to denote positive
quantities, the terms of the series on the right are
alternately positive and negative; also each term is less
than the preceding, for the ratio of the rth term to the
preceding term is *'^*'"* , which is less than unity since
^^=6^5',. +ar?r9» Hence the series, and thereforie the
continued fraction, is convergent provided the nth term
diminishes indefinitely when n is indefinitely increased.
It can be shewn that the condition of convergency is
satisfied whenever the ratio J^.^^j : a» is finite for all
values of n*
For let &,&,_, be always greater than A:.a„ where h is
some finite quantity.
Then u^ = ^^ =^^ = 1 8 J^i .
h h h
* 11^1 «— 1
Hence u < "'"n'"*^*' .
" !7»i?«.(l+A)
Whence u < i «
Todhunter's Algebra, Art. 783.
PERIODIC CONTINUED FRACTIONS. 461
But (1 + ky~* increases indefinitely with n, since k is
finite; hence u^ decreases without limit as n is increased
without limit.
We have therefore the following Theorem. The
in/mite continued fraction j^ IT tjT '"fi^ which ail the
letters represent positive quantities, is convergent if the ratio
K^n^i • ^n *^ always greater than some fixed finite quantity.
It should be remarked that any infinite continued
fraction of the form a + i  .... in which a, b, c,..,
+ c +
are positive integers, is convergent
362. In the following five Articles the continued
fractions will all be supposed to be of the form
^+ T t~ • • •» where a, 6, c,. . . are positive integers.
This form of continued fraction possesses two great
advantages, for we know that every convergent is in its
lowest terms, and we can also see by inspection, within
narrow Umits, the difference between any ^nvergent and
the true value.
•363. Theorem. Every simple periodic continued
fraction is a root of a quadratic equation with rational
coefficients whose roots are of contrary signs, one root being
greater and the other less than unity. Also the reciprocal
of the negative root is equal in magnitude to the continued
fra^ction which has the same quotients in inverse order.
Let the fraction be
11 1111
'^" ® ■*■ 6 + c + ••• + A: + f + a.+ 6 f •"•
Let ^ and ^ he the last two convergents of the first
* Articles 363, 364, and 368 are taken from a paper by Gerono,
Nouvelles Armales de Mathematiques, 1. 1.
462 PERIODIC CONTINUED FBACTION&
period. Then
a?Pf P'
.\a^Q + x(Q^P)^P'^0 (i).
The roots of (i) are obviously of different signs, and the
positive root is the value of the continued fraction.
Now, from Art 367, Ex. 1,
P'~^k+'" +b+a'
and _ = f + l^...^l.
Hence. ify = i + ^...^l^l^l^...,
we have ^^ ry + (^ ''
..y'P + y(QfP)Q = (ii).
The roots of (ii) are obviously of different signs, and the
positive root is the value of the continued fraction
1 11
From (i) and (ii) we see that the positive root of (ii) is
equal in magnitude to the reciprocal of the negative root
of (i); and therefore the reciprocal of the negative root of
<^^^{^+U+l+i+")'
The positive roots of (i) and (ii) are both greater than
unity, as may be seen by inspection; the negative root of
(i) must therefore be less than unity.
The fraction  , t . ••• . r . ~ ... requires no special
PERIODIC CONTINUED FRACTIONS. 463
examination, for we have only to change x into , and
X
y into ; thus  7 ••• . t t  •••is equal to the
positive root of Pa?— (Q—P) a? — Q = 0, and the negative
root is ■^f + 7 , ... , T . K
Hence, as before, one root of the quadratic equation in
X is greater and the other is less than unity.
364. Theorem. Every mixed periodic continued
Jraction, which has more them one nonperiodic element, is a
root of a qu/odratic equation vrith rational coefficients whose
roots are both of the same sign.
Let the fraction be
1 111 1111
and let
1 1111
A^ A
Let pT and ^ be the two last convergents of the non
periodic part ; then
yA + A' ,..
^ = ^BT£^ (^>
pf p
Let gr and ^ be the last two convergents of the first
period of y ; then
yP + F ....
^^yQ^Q
The elimination of y from (i) and (ii) will clearly lead
to a quadratic equation in x with rational coefficients.
Now, if the positive root of (ii) be substituted in (i) we
464 REDUCTION OF A QUADBATIC SURD
shall clearly obtain a positive value of x, and this will be
the actual value of the given continued firaction.
Also, from the preceding article, the negative value of
is — i^ + — . ... . !•; and, if this value be substituted
in (i), we have
_ ^1 11
we have to shew that this is positive. IS k>p the result
is obvious; ifk<v, =  ... is negative but is less
than 1, and therefore x is positive provided one element cU
least precedes k ; also k cannot be equal to v, for in that case
the periodic part would really begin with k and not with
OL Hence both values of x are positive in all cases.
Reduction of Quadratic Surds to Continued
Fractions.
365. It is clear that a quadratic surd cannot be equal
to a continued fraction with a finite number of elements ;
for every such continued fraction can be reduced to an
ordinary fraction whose numerator and denominator are
commensurable. It will be shewn that a quadratic surd
can be reduced to a periodic continued fraction of the form
a + r ,,..., where a, b, c, ... are positive integers.
The process will be seen from the following exampla
Ex. To reduce ^^8 to a continaed fraofdon.
The integer next below <^8 is 2; and we have
4
The integer next below "^—j — is 1 ; and we have
TO A CONTINUED FRACTION. 465
4 ""^+ 4 ^*4(V8+fl)^ V8+2
The integer next below ^8 + 2 is 4 ; and we have
^8+2=4+^82=4+ ^
' 78 + 2 *
4
The Btepa now recur, so that
,_ ^ 1 1 1 1
^®^+I+4+i+4+
••• •
Thus J& is equal to a periodic eontinned firaotion with one non
periodic dement, which is half the last quotient of the recurring
portion; and it will be proved later on that this law holds good
for every quadratic surd.
366. We now proceed to shew how to convert any
quadratic surd into a continued fraction.
Let /^N be any quadratic surd, and let a be the integer
next below ^/N; then
isJN + a \JIf\a '
whererj = JV— a*.
Since s/N—a is positive and less than 1, it follows
that is greater than 1. Let then h be the integer
next below — : then
 ^ _u_Zi:(^?l=.^)!_ . A ^ _i
where a^^br^^a and r, = — '■ ^
^t
n
Then, as before, * is greater than unity ; and if
s. A. 30
466 REDUCTION OF A QUADRATIC SURD
c be the integer next below ^ ^ , we have
N'(cr,a,y 1 •
where ctj = cr, — a, and r^ = '
^.
7*
'a
The process can be continued in this way to any extent
that may be desired. Thus JN= a + r .  . 3 .
•^ "^ b+ c+d +
367. To shew that any quadratic surd is equal to a re
curring continued fraction.
It is ftrst necessary to prove that the quantities which,
in the preceding Article, are called a, a,, a,,..., r,, r^, Tj,...
are all positive integers.
It is known that i\r is a positive integer, and that a, b,
c, d, ... are all positive integers.
We have the following relations :
r, = N a^ (i);
a^=br^a , r^r^ = Na^ (ii);
a3 = cr,a,, r^r^^Na^ (iii);
and so on.
Now it is obvious from (i) that r^ is an integer.
From (ii) we have r^ = ^^ — * — = 1 + 2a6 — 6V, ,
smce iY — a" = r, .
Thus a^=br^ — a, and r, =sl + 2a& — 6V, ; whence it
follows that a, and r^ are integers, since r^ is an integer.
TO A CONTINUED FRACTION. 467
From (iii) we have similaxly
aj«=cr, — a, and r^^r^h2a^c — <fr^;
whence it follows that a^ and r^ are integers, since a, and
r, are integers.
Then again, from (iv) we have
whence it follows that a^ and r^ are integers, since a^ and r^
are integers.
And so on; so that a« and r, are integral for all values
of n.
We have now to prove that a^ and r^ are positive for
all values of n.
We know that a, b, c, &c. are the positive integers
neoct below ^N, , i, &a Hence is/N—a,
r^ r,
hJN—a^y s/N—a^y &c., and therefore also N — a*, N—a^,
i\r— ttj*, &c., are all positive. That is r^, r,, r^, &c. are all
positive.
Again, since b is the integer next below , it
1
follows that ts/N^aKbr^ + r^. Now, a cannot be equal
or greater than ftrp for then /^Nkvi, and therefore a<r^;
therefore a<br^, since r, is positive and 6 a positive
integer. Hence a < 5rj, so that a, is positive.
Again, since c is the integer next below — *, it
follows that V^ + (i^<cr^ + V And we cannot have
«i><^t> for then \/N<r^, and therefore ciL<r,<crj, since
r, is positive and c a positive integer. Thus a^<cr^, so
that a, is positive.
And so on ; so that a^ is positive (or all values of n.
Having shewn that the quantities r^, r^, r,, &c. and a,
a,, ttj, &c., are all positive integers, it follows from the
30—2
468 REDUCmON OP QUADRATIC SURD.
. relation rjr^_j^ =^N — a^ that a. is less than VJV, so that
a,:J>a; hence the only possible values of a. are 1, 2,..., a.
Then, from the relation a, + a^+i = A?.r^, where A; is a
positive integer, it follows that r„ cannot be greater
than 2a.
Hence the expression ^ cannot have more than
2a X a different values ; and therefore after 2a* quotients,
at most, there must be a recurrence.
368. Theorem. Any qvadratic svrd can be reduced
to a periodic continued fraction with one nonrecurring
element, the last recurring quotient being twice the quotient
which does not recur; also the quotients of the recurring
period, exclusive of the last, are the same when read back
wards or forwards.
Let ts/N be the quadratic surd.
Then, from the preceding Article, we know that »JN
is equal to a periodic continued fraction.
We also know that any periodic continued fraction is
equal to one of the roots of a quadratic equation with
rational coefficients ; and the only quadratic equation in x
with rational coefficients of which i\/N is one root is the
equation a?* — JV = 0.
Now the roots of a? — N = are both greater than
unity in absolute magnitude, and the roots are o{. different
signs : it therefore follows from Articles 363 and 364 that
the continued fraction which is equal to ^/N must be a
mixed recumng continued fraction with one nonrecurring
element.
Hence we have
/AT ^1 1 1111
/AT 1 1 1111
SERIES EXPRESSED AS CONTINUED FRACTIONS. 469
Now T ..•••. r . r . T . r . •••is the positive
0+0+ +h+k+l+b+ ^
root of a quadratic equation with rational coefficients;
and as this positive root is /^N—a, the negative root
must be — V^— ct* Hence, Art 363, we have
1 1 1 1 1.1.1
:;/m'i+k+h+'" + cb'^i + ""
,^. _.7, 1 1 111
•'•'^^ ■^''"*"*"fc + A+  + 0+6 + 7 + ••••
u / ^1 1 111
Hence a + ^^jr_^ ... ^_^ ^ _^_^ ...
_ 1 1 1^1
"^■^6 + c++A + J+Z + "*'
whence it is easy to see that l^a^a, k = b, h = c, ....
Series expressed as Continued Fractions.
369. To shew that any series can be expressed as a
oontinued fraction.
Let the series be
t^i + Wj + W8 + w^+... +w„+ (i).
Then the sum of n terms of the series (i) is equal to
the nth convergent of the continued fraction
1 tt^ + W, — TZa + ^8 — '^+^4— *" — '^,^1 + W^ "
This can be proved by induction, as follows.
Assume that the sum of the first n terms of (i) is equal
to the nth convergent of (ii). Another term of the series
is taken into account by changing u^ into u^ + u^ ; and,
by changing t/^ into Un + ^nH* — *^ " will become
470 SKBIES BXPKESSED AS CONTINqia) FBACTION&
**"' ; ** — — , which is easily seen to be equid to
— *» * — *^ *^^ . Thus the sum of m + 1 terms . of
(i) will be equal to the (?i+l)th convergent of (ii) provided
the sum of n terms of (i) is equal to the nth convergent of
(ii). But it is easily seen that the theorem is true when
71 is 1 or 2 or 3: it is therefore true for all values of n.
Thus u^ + u^ + u^ + u^+ .,. to n terms
= ^ — ^ — — 7^ — ^* ... to w quotients... [A].
It can be proved in a precisely similar manner that
u^ — u^ + u^ — u^ + ^ ^ terms
= V . — ^— . — ^^ . — ^^ . ...to w quotients... [Bl.
The formula [B] can also be dedaoed from [A] by changing the
signs of the alternate terms.
370. The following cases are of special interest:
■P ± tM+ t t 1 ± ••• to ^ terms
6j 6,6, 6j6^
= T^ ^ , ^ ' ^ ir~ T ... to n quotients... [CI
all the upper signs, or all the lower signs, being taken.
1111
And — + — + — ±— +... to n terms
»! "" <*t ®8 ^4
= — _ —7 — ^ — 2 — .to w quotients ... [Dl
all the upper signs, or all the lower signs, being taken.
These can be proved by induction as in the preceding
Article.
SERIES EXPRESSED AS CONTINUED FRACTIONS. 47 1
Thus to prove [0]. It is obvious by inspeotion that the theorem
is true when n^2. Assume then that [C] is true for any particular
value of n; then, to include another term of the series ^ must be
changed into t^ ±\^tB , aiid therefore ^*~} will become
which can easily be seen to be equal to *~^ **  * **^^ — . Thus,
if [0] be true for any value of n, it will be true for the next greater
value ; hence as [C] is true when n=2, it is true for all values of n.
The following are particular cases of [C],
Oj + OjCb^ + aia^^ ± ajjOL^a/i^+
1.1 1 . 1
and — ± 1 ± h...
[E],
cbi a, a.
8
in*
aiTaa±lTa8±l + a4±l +
Ex. 1. Shew that
1+2 + 2 + 2+*^ "^^^*y
= r^5 + c~a + '«'*o iiiflnity. [Brounoker.]
1 9 u 7
Put ai=l, 03=8, a,s5, &Q, in [D].
Ex. 2. Shew that
1 1« 2* 8'
7 . T . 1 . T . — to infinity =log,2. [Buler.]
1+1+1+1+
Put 01=1, a,=2, 03=8, <&c. in [D].
* The formula [A] is due to Euler ; [0] is given by Dr Glaisher in
the Proceedings of the London Mathematical Society, Vol. v.
472 EXAMPLES.
Ez.8* Kndthetaloeofj j^ l^j^... toinfini^j.
Vtom [F] we iee thai
 i H £ U^iubutj
1 + 1 + 2 + 8+ "^ ^
13 3
Ex. 4* MndiheiiihcoiiTergeiitof n^^^^^..
From [F] we li«Te
1 1 1 1 ? ?
3"3T8"*'8.3.3" •■"3 + 2 + 2+ ••
Henee the nth convergent ieqmred= j ^ ~ ( "" s) I '
Ex.6. Shewihat l+j.fiirgirfsf+i "•=''^
 r r.r . r.r.r . r.r.r.r ^
^="^"*'i + 172"*" 17278 ■** 1.2.3 .4"*" •
EXAMPLES XXXYIL
1. Find the continued fractions equivalent to the following
quadratic surds :
(1) V17, (2) x/UO, (3) V33, (4) ^^3,
(5) V(^' + l), (6) V(a' + 24
2. Shew that V^= a + 2^ ^2^^..., where a has any value
whatever, and h^N^a\
3. Find the value of
EXAMPLES.
473
ttii) i  T I 1 H ...to infinity.
^ ' 2+3+4+5+4+5+
4. Shew that
5. Shew that
/I 1 1 1 1 Vd4 I i i ...)
\a + 6 +c +(i+a+ *"/ \ c+6+a+a+ /
b + d+bcd
" a + c + acb'
6. Shew that, if « = y + ^+^+  *^ i^^^^yi ^^"^
y^xTT TT ...to infinity.
^^ 2aj2aj
7. Shew that, if
1111 X . /: x
a; =  ^  r ... to infinity,
a + b +a+6 +
1111 X • fi •*
v = — ^ ^ HT ... to infinity,
^ 2a + 26 + 2a + 26 +
and » = ^ i^T^ia.*^"'^'^'*^^
3a + 36 + 3a + 36 +
then will aj(y»«') + 2y(«»fc') + 3«(»»y') = 0.
8. Shew thati if n be any positive integer,
, n»l» w"2» w»3»
'*=i+3~+~r"+"T~+
9. Shew that
l+a' + a*+... + a'" ,^_^1_ 1 ^
a + a' + a^+.. + a""* a ^_^1_^ . 1
_ •••
a+ — «+"
a a
to n quotients.
474 EXAMPLEa
10. Shew that, if
_a c a c , __c a c d
""6+5 + 6+5+"*^^ ^"5 + 6+5+6+ *"•
then bxdi/ = a — c»
11. Shew that the ratio of
1111 ,^1111
l+6+a+l+ l+a+6+l+
is 1 + a : 1 + 6.
12. Shew that the nth convergent of
14 2 2 . 21
13. Shew that the nth convergent of
1 1 1 (i^v2r(i^v2r'
'''*'2+2+2 + ^ (1 + ^2)" (1^2) •
12 2 2
14. Shew that the nth convergent of y « « ^
is2l.
15. Shew that the nth convergent of
1 ab db , oT "h*
... IS
ii+l_ A«+l'
a + 6 ~a + 6 a + 6— /* a"'*'^6
16. Find the nth convergent of the continued fraction
2 3 8 r^1
l.5_7^  2r+l
••• •
17. In the series of fractions — , — , <kc., where the law
of formation is J», = g^i, S'r = (»i* l)iE>^i + 2g^i; prove that
P 1
the limit of — when r is infinitely irreat is :; .
qr ® 1 + w
BXAMPL£S. 475
18. Shew that in the continued fraction
a, a, CL a.
6, + ft, + 63 + *" + ft, + '"'
the nth convergent is to the {n — l)th in the ratio
19. Shew that^ ^ ^p yg* <^m ^ the convergents of a simple
periodic continued fraction found by taking 1, 2, ^., complete
F F
periods, and if ^, ^y he the two convergents immediately
preceding y ^, then y^ = 'q ^''\ \ Q '
20. If Z be any integer not a perfect square, and if JZ
be converted into a continued fraction
11 1 Jl 1
ft + c + '** +A + 2a +ft "*'
and if the convergents obtained by taking one, two, ..., *
complete periods, each period terminating with k^ be denoted
by Pj, P„ ..., P„ prove that
F,^JZ^\F,^JZy
21. Find the nth convergent of the continued fraction
1 1 1
a + a~* — a + a~* — a + a~*  '" '
and shew that the limit of the nth convergent when n is
indefinitely increased is a or a"^ according as a is numerically
less than or greater than unity.
22. Shew that the nth convergent of
12 3 3 .3/ / 1\"\
2+2 + 2+2+'®8r V 3//
476 EXAMPLES.
2a Shew thai
 1 « X 2x (nl)x X • u j^
« =■= = Ts s ... ^ ^— ... to infinity.
24. Shew that
X 9? 9^ _X €US bx
a ab (we a+6— » + c — » +
25. Shew that
112 4 6 X ,: .^ J
T T o F ;? ••• to infinity = 6 '.
1+1+3+5+7+ ^
26. find the value of
113 6 3(n2) ^ . fi .^
Toko — o J" ... to infinity.
1+2+5 + 8+ + 3w4 + ^
27. Shew that
1 1V3^ 2'.4» 3'. 5' (nl)'(n+l)'
3+ 6 + 7 + 9 + ••• + 2w + l
1.3 2.4^3.5 •••^^ *^ n(n + 2)*
28. Shew that the nth convergent of
2 2 2 2 . ^ 2"l
T 77 V TT ... is 6
29. Shew that the nth convergent of
14 14 . 6nl + (l )"
3.3_3_3.. •••^«6n + 7 + (l)'"
2« 2' 3*
30. Shew that TT ^ tt ... to infinity =1.
3  5  7 
8L Shew that
,n 1 1 1.2 3.4 5.6 ..is«x
^2= 1 + 2 + r + T +3 +  *" "'*"*y'
EXAMPLES. 477
32. Shew that
 1 1 2.3 4.5 6.7 ^ . . .,
^~72 =2 + — + — +!" + •••*^"'^^'*y
33. Shew that
1 3.4 5.6 7.8 9.10
... to infinity = 1.
4_ 9  13  17  21 
34. Shew that (1 + x)*
_ ^ {nl)x 2(n2)aj 3(w3)a;
1 2 + (wl)aj3 + (7i2)a;4 + (r*3)a; •••
35. Shew that, if n be a positive integer,
91a.2 VLZl 2(n2) 3(»3) (wl)l
1— n + 1 n+1 — n+1   w+1
36. Shewthat(l+?)(l+J,)(H,)...
__  JB a (oj + a) a' (a? + a')
a x + a+a —x+a^a —
37. Shew that  75 = 5 ^ ... to infinity
n2n + l3n+l— "^
1 n 2n X * fi •*
CHAPTER XXVIIL
Theory of Numbebs.
371. Throughout the present chapter the word number
will always denote a positive whole number; also the word
divide will be used in its primitive meaning of division
without remainder. The symbol M(p) will often be used
instead of * a multiple ofp!
Definitions. A number which can only be divided
by itself and unity is called a prime number, or a prim£,
A number which admits of other divisors besides
itself and unity is called a composite number.
Two numbers which cannot both be divided by any
number, except unity, are said to be prim^ to one another^
and each is said to be prime to the other.
372. The Sieve of Eratosthenes. The different
prime numbers can be found in order by the following
method, called the Sieve of Eratosthenes,
Write down in order the natural numbers from 1 to
any extent that may be required : thus
1, 2, 3, 4, 6, 6, 7, §, d, lb
11, 12, 13, 14, 15, 16, 17, 18, 19, 20
2i, 22, 23, 24, 25, 26, 27, 28, 29, 30
31, 32, 33, 34, 35, 36, 37, 38, 39, 40 &c.
Now take the first prime number, 2, and over every
second number from 2 place a dot: we thus mark all
THEORY OF NUMBERS. 479
multiples of 2. Then, leaving 3 unmarked, place a dot
over every third number from 3 : we thus mark all mul
tiples of 3. The number next to 3 which is unmarked is
6 ; and leaving 6 unmarked, place a dot over every fifth
number from 5 : we thus mark all multiples of 5. And so
for multiples of 7, &c.
Having done this, all the numbers which are left
unmarked are primes, for no one of them is divisible by
any number smaller than itself, except unity.
It should be here remarked that if a composite number
be expressed as the product of two factors, one of these
must be less and the other greater than the square root of
the number, unless the number is a perfect square, in
which case each of the factors may be equal to the square
root. Hence every composite number is divisible by a
prime not greater than its square root. On this account
it is, for example, only necessary to reject as above mul
tiples of the primes 2, 3, 5 and 7 in order to obtain the
primes less than 121, for every composite number less than
121 is divisible by a prime less than 11.
373. Theorem. The nwmher of primes is infinite.
For, if the number of primes be not infinite, there
must l>e one particular prime which is greater than all
otheris. Let tnen p be the greatest of all the prime num
bers. Then \p will be divisible by p and by every prime
less than p. Hence Ip + 1 will not be divisible by p or
by any smaller prime; therefore lp + 1 is either divisible
by a prime greater than p, or it is itself a prime greater
than p. Thus there cannot be a greatest prime number ;
and therefore the number of primes must be infinite.
Ex. Find n donseoaiive nnmbers none of which i^re primes.
The nnmbers are given by [nfl + r, where r is any one of the
numbers 2, 8, ..., (n+1).
874. Theorem^ No rational integral algebraical
formula can represent prime numbers ordy.
480 THEORY OF im]fBKB&
For, if possible, let the ezpfessicm a±bx± ea^± daf±...
represent a prime namber for any integral valae of jc, and
for some particular constant intend valaes oi a^h,c^
Give to w any value, m suppose, such that the whole
expression is equal to j), where p is neither zero nor
unity ; then p = a±bm± cm* ± ... . Now give to ^ any
value m + Tip, where n is any positive int^ner ; then the
whole expression will be
a±b(m\np)±c(fn + np)* ± ... = a±bm±em* ± ...
+M(p)^p^M(p\
Thus an indefinite number of values can be given to x
for each of which the expression a±bx±ca?± ... is not
a prime.
In connexion with the above theorem, the following fonnnlae
ftre noteworthy : —
(i) a* 4 « + 41, whioh is prime if « < 40. [Enler.]
(ii) a* + « + 17, which is prime if « < 16. [Barlow.]
(iii) 2ir*+29,whichi8primeifd;<29. [Barlow.]
S76. The student is already acquainted from Arith
metic with many properties of factors of numbers : these
all depend upon tne following fundamental
Theorem : — If a number divide a product of two
factors, and he prime to one of the factors, it will divide
the other.
For, let ab be divisible by w, and let a be prime to x.
Reduce  to a continued fraction, and let  be the con
X q
vergent which immediately precedes  ; then [Art. 357, 1,]
qa —px = ± 1 ; .*. qah—pxh = ±6. Now qah is, by supposi
tion, divisible hy p\ and therefore qab—pxb must be
divisible by p, that is b must be divisible by p.
From the above theorem the following can easily be
deduced : —
THEORY OF NUMBERS. 481
I. If a prime number divide the product of several
factors it must divide one at least of the factors.
II. If a prime number divide a" it will divide a,
III. If a be prime to each of a, ^, 7, ... it will be
prime to the product afiy... .
IV. If a be prime to b, a* will be prime to 6"*.
V. If a number be divisible by several primes
separately it will be divisible by the product of them all.
376. Theorem. Every composite number can be re
solved into prime factors ; and this can be done in only one
way.
For, if JV be not a prime number, it can be divided
by some number, a suppose, which is neither JV nor 1:
thus JV= ah. Again, if a and b be not primes, we have
a^^oxd, b = e x/, and therefore JV= cdef. Proceeding
in this way, since the factors diminish at every stage,
we must at last come to numbers all of which are primes.
Thus JV can be expressed in the form ax^X7Xox...,
where a, ^8, 7, S, ... are all primes but are not necessarily
all different, so that JV may be expressed in the form
a')8*7*"..., where a, )8, 7,... are the different prime factors
ofiV.
Next, to shew that there is only one way in which a
number can be resolved into prime factors.
Suppose that JV = a6cd..., where a, b, c, d,,., are all
primes but are not necessarily all different ; suppose also
that JV = a^yS. . ., where a, )8, 7, S. . . are also primes. Then
we have abed... = afiyB... . Hence a divides a)S7S...; and
therefore, as all the letters represent prime numbers, a
must be the same as one of the factors of a)S7S.... Let
a^a; then we have bed... ^^^yS,.., from which it follows
that b must be equal to one or other of )8, 7, S, . . . ; and so
on. Hence the prime factors a, b, o,... must be the same
as the prime factors a, )8, 7, . . . .
Ex. Express 29645, 13689 and 90508 in terms of their prime factors.
Ans. 5 . V. 11^ 3M32 and 2^, 11«. 17.
S. A. 31
482 TDEORY OF NUMBERS.
377. To find the highest power of a prime number
contained in \n,
 J denote the integral part of  ; and let a be
any prime number. Then the factors in jn which will be
divisible by a are a, 2a, 3a, . . . , / f  j . a. Thus / f  j factors
of jn will be divisible by a. Similarly l(—%\ factors will
be divisible by a^ And so on.
Hence the whole number of times the prime number a
is contained in In is I [) "•'^("j "^■^("sj +••••
Ex. 1. Find the highest powers of 2 and 7 contained in [50.
If ^ J = 1. Hence 2*^ is the required highest power of 2.
Again, ^ { "S" ) = 7, I ( ^g ) = 1. Hence 7* is the required highest
power of 7.
Ex. 2. Find the highest powers of 3 and 5 which will divide 80.
Am. 33«, 6»
Ex. 8. Find the highest power of 7 which will divide [1000.
Am, 7i«.
378. Theorem. The product of any r consecutive
numbers is divisible by I r.
Let n be the first of the r consecutive numbers ; then
we have to shew that — ^^ — ^ ^ ^^^ ,
II
In + r — 1
or  . , , is an integer.
Jr ?i — 1 °
is n+ri^v, and the number of combinations of tif r — 1
THEORY OF NUMBERS. 483
things r together must be a whole number for all values
of n and of r.
The theorem can also be proved at once from first
principles by means of Art. 377.
For it is obvious that / ^^LtTzlj <t/ (^'^) + ^ £) '
/( 1 — j ^li — t'j^^iiji and so on. Hence from
Art. 377 it follows that the number of times any prime
number is contained in [nfr — 1 can never be less,
although it may be greater, than the number of times
the same prime number is contained in In — 1 x Ir. Thus
— 1^ Ir, occurs to
every prime number which occurs in
at least as high a power in Iw + r — 1, which proves that
\n + r'l is divisible by rt~ 1 x [r.
It can be proved in a similar manner that 
is an integer, where a + )S + 7+...=w.
379. If n he a prime number the coefficient of every
term in the expansion of (a + by except the first and last
terms is divisible by n.
For, excluding the first and last terms, any coefficient
, n(n— l)...(n — r + 1) ,
IS given by — —r^ ~ , where r is any integer
between and n.
Now, by the preceding Article, — ^^ ^^^
is an integer; and, as n is a prime number greater than r, n
must be prime to [r; and therefore ^^ — p — ^^
must be an integer. Hence every coefficient, except the
first and last, is cfivisible by n.
Similarly, if n be a prime number, the coefficient of
31—2
484 THEORY OF NXTMBEBa
every term in the expansion of (a + 6 + c+...)* which
contains more than one of the letters, is divisible by n.
'Eot the coefficient of any term which contains more
than one of the letters is of the form T—rkr > where
a+/8+7+...= n. Now . iy is an integer; and, as
w is a prime greater than any of the letters a, )8, 7, ..,
n must be prime to [al^lry... ; and therefore the coefficient
of every term which contains more than one letter is
divisible by n.
Ez. 1. Shew that n (n + 1) (2n + 1) is a multiple of 6.
Ex. 2. Shew that, if n be odd, (n^+S) (n«+7)=lf (32).
Ex. 8. Shew that, if n be odd, n^^M + ll=JJf (16).
Ex. 4. Shew that l + 7**+^=iJf (8).
Ex. 5. Shew that 19*»  1 = JJf (360).
Ex. 6. Shew that, if n be a prime number greater than 8,
n (n^  1) (na  4) = ilf (360).
380. Fermat's Theorem. If n he a prime number,
and m any number prime to n; then m""* — 1 will he
divisible by n.
We know that when n is a prime number, the coeffi
cient of every term in the expansion of (a^ + a^ +...+ a^)",
which contains more than one of the letters, is divisible by
n. Now there are m terms each of which contains only
one letter and the coefficient of each of these terms is 1.
Hence, putting a^ = a, =...= 1, we have
m''==m + M{n) ; /. m (m*** — 1) = M{n).
Hence, if m be prime to ti, m""* — 1 will be a multiple
of w.
Ex. 1. Shew that, if n be a prime nmnber,
li*i + 2*i + 8»i+... + (nl)»i+l = JJf(n).
Ex. 2. Shew that, if a and h are both prime to the prime nmnber n\
then will a*"^  6*^^ be a multiple of n.
Ex. 3. Shew that n»  n= If (30).
THEORY OF NUMBERS. 485
Ex. 4. Shew that n7  n=lf (42).
Ex. 5. Shew that x^^  2/^=^(1365), if a; and y are prime to 1365.
Ex. 6. Shew that, if m and n are primes; then
Ex. ?• Shew that, if m, n and p are all primes; then
(n))"*i + (pm)^^^ + (mny*"^  1 = Jlf (mnp),
Ex. 8. Shew that the 4th power of any number is of the form 5m
or 5m+l.
Ex. 9. Shew that the 12th power of any number is of the form 13ni
or 13m +1.
Ex. 10. Shew that the 8th power of any number is of the form 17m
or 17m ±1.
381. To find the number of divisors of a given number.
Let the given number, N, expressed in prime factors,
be a*6V Then it is clear that N is divisible by
every term of the continued product
(l + a + a'+...Ha')(l + 6 + 6*+...+ 6^)(l + c + c'H...Hc*)...
Hence the number of divisors of JV, including N and
(^ + l)(y + l)(^ + l)
Ex. 1. The number of divisors of 600, that is of 2^ . 8 . 5^, is
(8+l)(l + l)(2 + l)=24.
Ex. 2. Find the sum of the divisors of a given number.
The given number being N=a*bV(f... , the sum required is easily
seen to be
(1  a^^) (1  5y+i) (1 _ c^i)...
(la)(l6)(lc)...
Ex. 8. Find the number of divisors of 1000, 8600 and 14558.
Am. 16,45,24.
Ex. 4. Shew that 6, 28 and 496 are perfect numbers. [A perfect
number is one which is equal to the sum of all its divisors, not
considering the number itself as a divisor.]
Ex. 5. Find the least number which has 6 divisors. Ans. 12.
Ex. 6. Find the least number which has 15 divisors. Ans. 144.
Ex. 7. Find the least number which has 20 divisors. Ans. 240.
Ex. 8. Find the least numbers by which 4725 must be multiplied in
order that the product may be (i) a square, and (ii) a cube.
Ans. 21, 245.
486 THEORY OF NUMBERa
382. To find the number of pairs of factors, prime to
each other, of a given number.
Let the given number be J\r= a%^c*.,, ; then, if one of
two factors prime to each other contains a, the other does
not ; and so for all the other different prime factors.
Hence the factors in question are the different terms
in the product (1 + a*) (1 + 6*') (1 +c*)..., the number of
them being 2", where n is the number of different prime
factors of N, The number of different pairs of factors
prime to each other is therefore 2""^ in which result N
and 1 are considered as one pair.
383. To find the number of positive integers which are
less than a given nu/mber and prime to it.
Let the given number be JV"= a'b^c'.,., where a, 6, c,...
are the different prime factors of N,
The terms of the series 1, 2, 3,..., iV which are divisible
by a are a, 2a, 3a,..., — a; and therefore there are —
•^ a ' a
N
numbers which are divisible by a. So also there are j
iV . . . JV . . .
numbers divisible by 6, j divisible by be, y divisible by
abc, and so on.
We will now shew that every integer which is less
than i\r and not prime to JV" is counted once and once only
in the series
^N ^ir ^^ N ^ If ^ , ,
a ab abc abed
Suppose an integer is divisible by only one prime factor
of N, a suppose; then that integer is counted once in
N
(a), namely as one of the — numbers which are divisible
by a.
Next suppose an integer is divisible by r of the prime
fieictors a, b, c,... , then that integer will be counted r
THEORY OF NUMBERS. 487
times in S — , it will be counted \ _  times in 2 7 >
Of 1.2 (w
it will be counted — ^^ — = — ^^r times in S 7 , and so
1.2.0 aoc
on. Hence the whole number of times an integer divisible
by r of the prime factors is counted, is
r(r~l) r(r~l)(r2) ., ,,^, r(r~ l),..l
^ 1.2 ■*■ 1.2.3 •••■^^ ^ [^
= l.(liy = l.
Thus every number not prime to iV is counted once in
(a); and therefore the number of positive integers less
than N and not prime to iV is given by (a) ; provided
however that unity is considered to be prime to N.
Hence the number of positive integers less than JV and
prime to N is
iVS — 12^27 + •••
a ao aoc
[ a <w aoc )
= ^(19(15) (19 f^^^^J
Ex. 1. Find the number of integers less than 100 and prime to it.
Since 100=2^ . 5^ the number required is
100 (li)(li) 1=39.
Ex. 2. Find the number of integers less than 1575 and prime to it.
Am, 719.
Ex. 3. Shew that the number of integers, including unity, which
are less than 2^[7^>2] and prime to N is even, and that half
N
these numbers are less than ^ .
r N
For if a be prime to 2/ so also is Na; and if a > , then
488 THEORY OF NUMBERS.
384. Forms of square numbers. Some of the
different possible and impossible forms of square numbers
will be seen from the following examples : —
Ex. 1. Shew that every equare is of the fonn 3ni or 3ni + 1.
For every number is of the form 8m or 8m sfts 1. Hence every
square is of the form 9m or 3m+ 1«
Ex. 2. Shew that every square is of the form 5moi 5mdkl,
For every number is of the form 5m, 5m ± 1 or 5m ± 2 ; and there
fore every square is of the form 5m, 5m +1 or 5m +4.
Ex. 8. Shew that, if a^ + 5^= c^, where a, b, c are integers ; then will dbe
be a multiple of 60.
First, every square is of the form 8m or 8m+l; and therefore
the sum of two squares neither of which is a multiple of 8 is of the
form 8m +2 which cannot be a square. Hence either a or b must be
a multiple of 3.
Again, every square is of the form 5m or 5m ± 1. The sum of two
squares neither of which is a multiple of 5 is therefore of one of the
forms 5m, or 5m db 2. Now no square can be of the form 5m :^ 2 ; and
if a square be of the form 5m, its root must be a multiple of 5.
Hence, if ab is not a multiple of 5, c will be a multiple of 5. Thus,
in any case, abc is a multiple of 5.
Lastly, since every number is of the form 4m, 4m +1, 4m +2 or
4m +8, every square is of the form 16m, 8m + 1, 16m +4. Now a
and b cannot both be odd, for the sum of their squares would then be
of the form 8m + 2 which cannot be a square. Also, if one is even
and the other odd, the even number must be divisible by 4, for the
sum of two squares of the forms 8m +1 and 16m +4 respectively is of
the form 8m + 5 which cannot be a square. It therefore follows that
ab must be a multiple of 4.
Thus abc is divisible by 8, by 5 and by 4; hence, as 8, 4 and
5 are prime to one another, abc=M(QO).
Ex. 4. Shew that every cube is of the form 7m or 7m ± 1. Shew also
that every cube is of the form 9m or 9m ds 1.
Ex. 5. Shew that every fourth power is of the form 5m or 6m + 1.
Ex. 6. Shew that no square number ends with 2, 8, 7 or 8.
Ex. 7. Shew that, if a square terminate with an odd digit, the last
figure but one will be even.
Ex. 8. Shew that the last digit of any number is the same as the last
digit of its (4n+ l)th power.
Ex. 9. Shew that the product of four consecutive numbers cannot be
a square.
THEORY OF NUMBERS. 489
EXAMPLES XXXYIIL
1. Shew that the difference of the squares of any two
prime numbers greater than 3 is divisible by 24.
2. Shew that) if n be a prime greater than 3,
w(w« l)(w« 4)(w*  9) = if (2^ 3'. 5 . 7).
3. Shew that, if » be any odd number,
(n + 2m)"  (rn 2m) = if (24).
4. Shew that a*+'  a*"*' = if (30).
5. Shew that) if iT— a* = a5 and (a+1)* — iV = y, where x
and y are positive; then iVscy is a square.
6. How many numbers are there less than 1000 which are
not divisible by 2, 3 or 5?
7. P, Qf By p, q^ r are integers, and jp, g, r are prime
P O R
to one another; prove that, if— h — + — be an integer, then
— , — and — will all be integers.
p q r
8. Shew that 284 and 220 are two 'amicable' numbers,
that is two numbers such that each is equal to the sum of the
divisors of the other.
9. Shew that, if 2"l be a prime number, then 2""*(2"l)
will be a 'perfect* number, that is a number which is equal to
the sum of its divisors.
10. Find all the integral values of x less than 20 which
make «'•  1 divisible by 680.
11. Shew that no number the sum of whose digits is 15 can
be either a perfect square or a perfect cube.
12. Shew that every square can be expressed as the differ
ence between two squares.
490 THEORY OF NUMBERS.
13. Find a general formula for all the numbers which when
divided by 7, 8, 9 will leave remainders 1, 2, 3 respectively;
and shew that 498 is the least of them.
14. If n be a prime number, and iT prime to n, shew that
iVr»*i»_l = Jf (n"), and that iVT*'*'"* 1 = Jf (n').
15. Shew that, if to be a prime number and iV be prime to
n, then wiU iyri+«+ +<« ^ l =M{n').
16. Shew that, if /? be a prime number, and (1 + xy= 1 +
a^x\'a^af + a^x'h,..; then a, + 2, a, 3, a^^i, <kc. will be
multiples of p,
17. Shew that if three prime numbers be in A. p. their
common difference will be a multiple of 6, unless 3 be one of
the primes.
1 2a (26
18. Shew that , ., , , is an integer.
[al6a + 6 ^
2n
19. Shew that . — =t is an integer.
[n+ 1 \n
\nr
20. Shew that , TTi* ^^ an integer.
\n{\r}
21. Each of two numbers is the sum of n squares ; shew
that the product of the two numbers can be expressed as the
sum of ^ TO (to— 1) + 1 squares.
22. Shew that a' + 6* cannot be divisible by 3, unless both
a and b are divisible by 3; shew also that the same result holds
good for the numbers 7 and 11.
23. Shew that, if a' + b' = c\ then ab{a'b*) wiU be a
multiple of 84.
24 Shew that no rational values of a, b, c, d can be found
which will satisfy either of the relations a' + 6' = 3 (c* + rf"),
a' + b' = 7 {c' + d') or a' + b':=n {c'\'d%
25. Shew that, if a* + c" = 26», then a'b' = M (24).
THEORY OF NUMBERS. 491
Congruences.
385. Definition. If two numbers a and h leave the
same remainder when divided by a third number c, they
are said to be congruent with respect to the modulus c; and
this is expressed by the notation a = 6 (mod. c), which
is called a congruence.
For example, 21 = 1 (mod. 10), and (a +1)^=1 (mod. a).
The congruence a=h (mod. c) shews that a — 6 is a
multiple of c, which can be expressed by
a — 6 = (mod. c).
386. Theorem. Ifa^ = h^ {mod, x), and a^ = b^(mod. x) ;
then will a^\a^ = \ + b^ (mod, x), and afl^ = hp^ {mod, x).
For let a^ = m^x + r , and a, = mjjc + r, ; then, by sup
position, 6j = n^a? + r^ and 6, = n^a? + r,.
Hence a^ + a, — (6^ + h^ = (m^ + m^ — n, — n^) a? ;
.. {a^ + ctj)  (6j + 6j) = (mod. a?),
or a^ + ttj = 6j + 63 (mod. x).
Again, it is easily seen that afl^ — 6^6^ = a multiple of
X, and therefore a^a, = 6^6, (mod. a;).
The proposition will clearly hold good for any number
of congruences to the same modulus.
387. Congruences have many properties analogous to
equations. For example, if the congruence
Aa? + Bx\G=0{mo^p)y
wherein A, B, C have constant integral values, be satisfied
by the three values a, b, c of x, which are such that a — b
is unity or prime to p, and so for every other pair, then
the congruence will hold good for all integral values of x,
and A, B, C will all be multiples of p.
For we have
Aa"" + Ba + O^0 (mod. p\
and Ab'\Bb + G = (mod. p) ;
492 THEORY OF NUMBERS.
/. by subtraction
(a b){A(a + b) + B}=0 (mod. p).
Hence, as a — 6 is unity or prime to jp, we have
A(a + b) + B = (mod. p).
Similarly, A{a + c) + B=0 (mod. p).
Hence, by subtraction, A {b — c) = (mod. p).
Therefore A = (mod. p), since 6 — c is unity or prime
to p.
Then, since A = (mod. p), it follows that jB = (mod.
p), and then that = (mod. p).
Then, since A, B, G are all multiples of p, it follows
that Aa^ + Bx + (7 is also a multiple of p for all integral
values of x.
We can prove in a similar manner the general theorem,
namely : —
If a congruence of the nth degree in x be satisfied by
more than n values of x, which are such that the difference
between a/ny two is unity or is prime to the modulus, then
the congruence will be satisfied for all integral values of x,
and the coefficients of all the different powers of x mil be
multiples of the modulus,
388. Theorem. If a and b are prime to one a/nother,
the numbers a, 2a, 3a,..., (6 — 1) a will all leave different
remainders when divided by b.
For suppose that ra and sa leave the same remainder
when divided by b.
Then ra sa= M(b); but if b divide (r — s) a, and be
prime to a, it must divide r — 5, which is impossible if r
and s are both less than b.
Hence the remainders obtained by dividing a, 2a, ... ,
(6  1) a by 6 are all diflferent; and since there are 6 —1 of
these remainders, they must be the numbers 1, 2, ,
(6 — 1) in some order or other.
THEORY OF NUMBERS. 498
If a be not prime to h the remainders obtained by dividing a, 2a,
3a, ..., (b1) ahj b will not be all different. For let A; be a common
factor of a and b, and let a = ka and & =s kp. Then it is easily seen
that (r +/3) a and ra will leave the same remainder when divided by 6,
and (r+p)a and ra are both indnded in the sreies a, 2a, ..., (6  1) a
provided r + j8 < 6  1,
Cor. If a be prime to 6, and n be any integer what
ever, the remainders obtained by dividing n, n +a,
n+2a, ... , n + (b — 1) a by 6 will all be different, and will
therefore be the numbers 0, 1, 2, ... , (6 —1).
389. Fermat's Theorem. From the result of the
preceding article, Fermat's theorem can be easily deduced.
For, if a and b are prime to each other, the numbers a,
2a, ..., (6 — 1) a will leave, in some order or other, the re
mainders 1, 2, . . ., (6 — 1), when divided by 6. Hence we have
a.2a. 3a... (6 — l.a) = 1 .2.3...(6 l)(mod. b).
that is 61 (a^^  1) = (mod. b).
Now, if 6 be a prime number, 16—1 will be prime to 6;
and we have a^^ — 1 = (mod. 6), which is Fermat's
theorem.
390. Wilson's Theorem. If n be a prime nvmber,
1 + 1^ — 1 will be divisible by n.
If a be any number less than the prime number
n, a will be prime to n, and hence, from Art. 388,
the remainders obtained by dividing a, 2a, ..., (w — 1) a by
n will be the numbers 1, 2, ..., (n — 1); hence one and
only one of the remainders will be unity. Let then db be
the multiple of a which gives rise to the remainder 1; then,
if b were equal to a, we should have a^^l + M(n), or
(a + 1) (a — 1) = M{n)y and this can only be the case, since
n is a prime, if a = l or a = 7i — 1. Hence the numbers
2, 3, ... (rt  3), {n — 2) can be taken in pairs in such a way
that the product of each pair, and therefore the product of
all the pairs, is of the form M{n) + 1.
Thus 2 . 3 . 4...(n  2) = M\n) + 1 ;
.. In  1 = M{n) X (n  1) + n  1.
494 THEORY OP NUMBERS.
Hence I n — 1 4 1 = M{n).
Wilson's theorem may also be proved as follows : —
From Art. 305, we have
^ ^ w — 2 '
Now by Format's theorem (n  1)""* = 1 + M{n),
{n  2)»^ = 1 + Jf (n), &c.
Hence we have
1 _ (n _ 1) + (iLnlI^Jl!) _... + (_ l).Hi („_ 1) + Jlf(„)
w — 1,
that is (1  1)"'  ( ir'+if (n) = \nl \ hence, as w 1
is even, I n — 1 4 1 = M(n),
Wilson's theorem is important on account of its express
ing a distinctive property of prime numbers; for 1 + [^—1
is not divisible by n unless n is a prime. For if any
number less than n divide n it will divide jw — 1 and
therefore cannot divide n — 1 + 1.
391. Theorem. If the nwmher of integers less than
any number n and prime to n be denoted by ^ (n) ; then, if
a,b,c,,., are prime to each other,
<l>{ahc,..) = <l>{a) X <^(6) x <^(c) ,
provided that unity is considered to be prime to cmy greater
number.
First take the case of two numbers a, b and their
product ah.
Arrange the ah numbers as under :
1 , 2 , 3 , , k , a
a + 1 , al2, a + 3, , a + k, 2a
2a+l, 2af2, 2a+3, , 2a + k, Sa
(6l)a+l, (6l)a+2, (6l)aro, ..., (6l)a+/j, ... ba.
THEORY OF NUMBERS. 495
Then it is clear that all the integers in the kth. vertical
column will or will not be prime to a according as A; is or
is not prime to a. Hence there are (f) (a) columns of
integers, including the first; all of which are prime to a.
Then again, we know from Art. 388 that since a is prime
to b, the remainders obtained by dividing the numbers
k, a + k,.,.y (b — 1) a + k by 6 are the numbers 0, 1, 2,...
(6 — 1); and it is clear that a number is or is not prime to
b according as the remainder obtained by dividing the
number by b is or is not prime to b. Hence there are
as many integers prime to 6 in any one column as there
are in the series 0, 1, 2,... (6 — 1), that is to say, there are
in each column (f>{b) integers prime to b. Thus there are
<f) (a) columns of integers prime to a and each column
contains <f> (b) integers which are also prime to b. But all
integers which are prime to a and also to b are prime to
axb. Hence the number of integers less than ab and prime
to oi is ^ (a) X <f> (b), so that <f> {ab) = <f>(a)x<f> (b).
The proposition can at once be extended, for we have
(f>(ahc...)= <f> {axbc.) = (f> {a) x<j)(bc,,,)
= <f>{a)<f>(b)<f>{c..,)
= <l>(a).<f>{b).<f){c)...
. 392. The number of integers less than a given number
and prime to it can be found by means of the theorem in
the preceding article.
For let the number be iV=a"6^c>'..., where a, 6, c,...
are the dififerent prime factors of N.
To find the number of integers less than a" and prime
to it, (unity being considered as one of these numbers) we
must subtract a**~^ from a*; for the numbers a, 2a, 3a,...,
a*^ . a are not prime to a, and these are the only numbers
which are not prime to a ; thus
<f> (a*) = a*  a*i = a* ( 1   V
496 THEORY OF NUHBEBS.
Similarly ^{¥) = b^(lh , <!> (c>) = ct (l  J) , &c.
But, by the preceding article,.
= „.(l_l).^(l_l).,v(l_J)....
Hence <^ W = jr(l  ^) (l 1) (l J)....
where a, 6, c,... are the different prime factors of N, and
unity is considered to be prime to a, 6, c, &c.
393. The following is an extension of Fermat's
Theorem : —
If a and m are two numbers prime to one another,
and <j> (m) the number of integers, including unity, which
are less tha/n m and prime to m; then a*w — 1 =
(mod, m).
Let the ^ (m) integers less than m and prime to m be
denoted by 1, a, )8, 7,. . . , (m — 1). Then the products a . 1,
aa, a^, arf,, . ., a (m — 1) must all leave different remainders
when divided by m, for if any two, ra and sa suppose, left
the same remainder, (r — 5) a would be a multiple of m,
which is impossible since a is prime to m and r — « is less
than m. Moreover the remainders must all be prime to
m, since the two factors of any one of the products are both
prime to m; and therefore as the ^(m) remainders are
all different, and are all prime to m, they must be, in some
order or other, the ^(m) numbers 1, a, )8, 7...
Hence
a.aa.ajS a(m — 1) = 1 .a .13 .7... (m — l)(mod. m);
/. {a*(^>.l}l.a./8...(ml)=0(mod. m).
Hence as 1. a./8...(m — 1) is prime to m, we have
a^im) — 1 = (mod. m).
If m be a prime number, <j> (m) = m — 1, and we have
Format's Theorem.
J
THEORY OF NUMBERS. 497
3.94. Lagrange's Theorem. Ifp he a prime number,
the sum of all the products r together of the numbers 1, 2, 3.
..., jp — 1, is divisible by p^r being any integer not greater
than p — 2.
Consider the identity
{x'l)(x2),..{cop + l) = x^'S,x'^ + S^i€^'
+ ...+ (17^5^.,.
Change x into x — l; then
{x2)(xS)...(x'p)^(xiy'S,{xiy^+,..
Hence
Equate the coeflScients of the different powers of a; in
the above identity; and we have
1.0, J 2 ,
,, _ p(pl)(p2) „ (pl)(p2)
^^' 1.2.3 "^ '• 1.2
o q  J'(j>l)(p2)(p3) , o (pl)(p2)(p3)
^ »~ 1.2.3.4 '• 1.2.3
(p2)(y3)
** 12 '
r„ 2^ « _ p(pl)2 g (pl)(p2 )...2
y> ^>''^»»i.2...(^l)"^^> 1.2...(p2)
(p 2)...2 3^
^ » 1.2...(p3)^***^ '^•1.2*
Since J) is a prime the first term in each righthand
member is divisible by p ; whence it follows from the first
equation that 8^ is a multiple of p, and then that /S, is a
mtiltiple of p, and so on to iS^,.
s. A. 32
498 THEORY OF NUMBERS.
Lagrange's Theorem may also be deduced from the
Theorem of Art. 387, assuming that Fermat's Theorem is
known.
For the congruence
{x 1) {x  2).. .(a? jp + 1) a?*"' + 1 = (mod. p\
is of the (n — 2)th degree in a?, and by Format's Theorem
it is satisfied by the p—1 values 1, 2, ..., jp — 1, which are
such that the diflference between any pair is either unity
or is prime to p. Hence, by Art. 387 it is true for all
integral values of a?, and the coeflScients of all the dififerent
powers of x are multiples of jp.
It should be remarked that Wilson's Theorem follows
at once by putting a? = 0.
395. Reduction of firactions to circulating^
decimals.
It is obvious that a fraction whose denominator con
tains only the factors 2 and 5 can be reduced to a ter
minating decimal, for
a _ a.5^.2«
2'5* ~ 10^ '
If, however, the denominator contains any foctor which
is prime to 10, the fraction can only be reduced to a cir
culating decimal.
Let the fraction in its lowest terms be y  , where
h is prime to 10. Let this fraction be equivalent to a cir
culating decimal with a recurring and ^ nonrecurring
figures.
Then
a _ a.5''.2^ ^ N ^
2" . 5^ 6 " lQ^^\h " 10^ (lO'^ 1) '
/. lO'^.6.i^^=a.5^2^lo^(lo«l).
Hence, as 6 is prime to a and to 10, 10* — 1 = Jf(fe),
cmd OL is the lowest power of 10 for which this is true, for
i
THEORY OF NUMBERS. 499
otherwise the fraction could be expressed as a circulating
decimal with fewer than a recurring figures.
It should be noticed that the number of recurring
figures in the circulating decimal depends only on b and is
not affected by the presence of 2''6* in the denominator,
for the number is a, where a is the lowest power of 10
which is equal to if (6) + 1.
We will now prove that a is either equal to ^ (6) or to
one of its submultiples.
By the extension of Format's Theorem [Art. 393] we
have
10*<*)1=: if (6).
We have also 10*  1 = M{h).
Hence, if a be not ^ (6) or one of its submultiples, let
if> (b) = Aa h r, where r < a.
Then 10*<*)  1 = 10* . lO*"  1
= {if (6) + IIMO*  1 = if (6) h W  1 ;
10'"l = if(6),
which is impossible since r<a, and a is the lowest power
of 10 which is equal to if (6) h 1.
Hence, if b be the factor of the denominator of a fraction
which is prime to 10, the number of recurring figures in the
equivalent dedmalis either <j> (b) or one of its submultiples,
396. We shall conclude this chapter by considering
the foUowing examples :—
Ex. 1. Shew that 3*»+2  8n  9 is a multiple of 64.
We have
32»+«8n9=(l + 8)'Hi_8n9=l + (n+l)8 + ilf(82)8n9=ilf(8»).
Ex. 2. Shew that 3«*  32n« + 24n  1 = (mod. 512).
Let u„=32»32n«+24?il;
then t4^i=3**+»32(n+l)»+24(n+l)l.
Hence w^i9u^=266n»266n=256n(nl)=M(512),
since n (n  1) is divisible by 2.
32—2
500 EXAMPLES.
And since WH+i9ti,^ = (mod. 512), it follows that w«+.i^O
(mod. 512) provided u^^^O (mod. 512). The theorem is therefore
true for all values of n provided it is true for ns=l, which is the case
since ti^sO.
Ex. 3. Shew that no prime factor of n'+ 1 can be of the form 4m— 1.
Every prime nmnber, except 2, is of the form 2k +1. Let then
2X; + 1 he a prime factor of n^+1. Then n is prime to 2k + 1, aivd
therefore by Fermat's theoFcm n^=M {2k + 1) + 1.
But, by supposition, n* + 1 = 3f {2k + 1) ;
n^={M{2k+l)l}*=M{2k + l) + {'l)K
Since n»=M {2k + l)hl and n»=Jlf (2A; + !) + ( 1)» it foUows
that k must be even, and therefore every prime factor of n'+l is of
the form 4?n+l, and therefore no prime factor can be of the form
4i»l.
Since the product of any number of factors of the form 4m + 1 is
of the same form, it follows that every odd divisor of n^+1 is of the
form 4m +1.
Ex. 4. Shew that every whole number is a divisor of a series of nines
followed by zeros.
Divide the successive powers of 10 by the number, n suppose, then
there can only be n different remainders including zero, and hence
any particular remainder must recur. Let then 10* and lO*' leave
the same remainder when divided by n : then 10*  lO*' is divisible by n
and is of the required form.
EXAMPLES XXXIX
1. Prove the following : —
(i) 2*»+*  9ri« + 3to  2 = if (54).
(ii) 5*+* + 7i*5ri» + 4w5 = Jf(12p).
(iii) 4''» + 3+« = 0(mod. 13).
(iv) 3*"+» + 2 . 4'^"*^ = (mod. 17).
2. Shew that, if a be a prime number, and b be prime to
a; then IV, 2"6«, f^LzlYh' will give different re
mainders when divided by a.
3. Shew that, if 4w + 1 be a prime number, it will be a
factor of { 1 2n}* + 1 ; and that, if 4w — 1 be a prime, it will be a
factor of {[2nl}«l.
4. Shew that, if w be a prime number, and r be less than
n; then will Ir — 1 In — r + (— l)'""^ = i/(n).
EXAMPLES. 501
5. Shew that^ if m and n are prime to one another, every
odd divisor of m* + n' is of "the form 4^ + 1.
6. Shew that T:+s= + :r: + 7^ + «'« to infinity
where 2, 3, 5,... are the prime numbers in order.
7. Shew that the arithmetic mean of all numbers less than
n and prime to it (including unity) is ^.
8. Shew that, if iV be any number, and a, &, c, ... be its
different prime factors ; then the sum of all the numbers less
than N and prime to iV is ^ ( 1 — ] (1 — r) (l — ) •••, and
the sum of the squares of all such numbers is
x(iS(i^)"4(\»)(i*)
9. If tj> (m) denote the number of integers less than m
and prime to it; and if c?^, d^^ c^.v ^® ^^ different divisors
of n'y then will '%tj>(d) = n,
10. Shew that, if a fraction ^ , where h is prime and prime
to 10, be reduced to a decimal, and if the number of figures
in the recurring period is even ; then the sum of the first half of
the figures added to the last half will consist wholly of nines.
11. If  be converted to a circulating decimal with p — \
It
figures in its recurring period, shew that p must be prime and
that the recurring period being multiplied by 2, 3, (p — 1)
will reproduce its own digits in the same order.
12. Shew that, if ^ has a circulating period oi p figures, ^
of q figures, and pOir figures,..., and if P, Q, By,,, are prime,
then pTTn — ^11 have a circulating period of n figures, where
n is the l.c. m. of ^, q, r....
CHAPTER XXIX.
Indeterminate Equations.
397. We have already seen that a single equation
with more than one unknown quantity, or n equations
with more than n unknown quantities, can be satisfied in
an indefinite number of ways, provided there is no restric
tion on the values which the unknown quantities may
have. If, however, thq values of the unknown quantities
are subject to any restriction, n equations may suflBce to
determine the values of more than n unknown quantities.
We shall in the present chapter consider some cases of
equations in which the unknown quantities are restricted
to integral values.
398. It is clear that every equation of the first degree
with two unknown quantities x and y can be reduced
to one or other of the forms ax + by=±c, ax — by^ ±Cy
where a, b, c axe positive integers.
By changing x into —x and y into — y, ax + by = c
will become ax + by = c, and ax — by^c will become
— cw? + 6y = c ; hence in order to shew how to find
integral solutions of any equation of the first degree in x
and y, it is only necessary to consider the two types
dx + by^o and ax — by^c.
Now, it is evident that the equation ax±by=^c cannot
be satisfied by integral values of x and y, if a and b have
any common factor which is not also a factor of c ; and, if
a, 6 and c have any common factor, the equation can be
INDETERMINATE EQUATIONS. 603
divided throughout by that factor. In what follows it
will therefore be supposed that a and b are prime to one
another.
399. To shew that integral values can always he found
which will satisfy the equation a^±by = c, provided a and
b are prime to one another.
Let T be reduced to a continued fraction, and let
 be the convergent immediately preceding r. Then,
from Art. 367,
.*. a(±cq) — b(±qp) = c (i),
and a{±cq)'{b{+cp) = c (ii).
Hence it follows from (i) that either x^cq, y — cp or
a? = — cj, y = — cp is a solution of the equation ax — by=^c;
and from (ii) that either x = cq, y = 'cp or a? = — eg,
y = cp is a solution of the equation ax + by ==c.
Hence at least one set of integral values of x and y can
always be found which will satisfy the equation ax ±by=c.
The above investigation fails when a or 6 is unity.
But the equation ax ±y = c is obviously satisfied by the
values 35 = a, ±y = c — oa, where a is any integer. So also
a? + 6y = c is satisfied by the values a? = c + 6)8, y=^l3, where
)8 is any integer.
Hence the equation ax±by = c always admits of at
least one set of integral values.
400. Having given one set of integral values which
satisfy the equation ax — by = Cf to find all other possible
integral solutions.
Let x = a, y = fi be one solution of the equation
ax — by=c; then aa — b^ = c. Hence, by subtraction,
a(^a)6(y/8) = 0.
I
604 INDETERMINATE EQUATIONS.
Now since a divides a (a? — a), it must also divide
6 (y — /9) ; a must therefore be a fiEtctor of y — /8, since it is
prime to 6.
Let then y — /8 = ma, where m is any integer ; then
a (a? — a) = mta, and therefore a? = a + mb.
Hence, if a; = a, y = 13 be one solution in integers of
the equation aa? — 6y = c, all other solutions are given by
a? = a + mh, y = /8 + ma,
where m is any integer.
It is clear from the above that there are an indefinite
number of sets of integral values which satisfy the equation
aa;—by = c, provided there is one such set ; and, from the
preceding article, we know that there is one set of integral
values.
It is also clear that, whether a and /8 are positive
or not, an indefinite number of values can be given to rn,
which will make a + mb and /8 + ma both, positive.
Hence there are an infinite number o{ positive integral
solutions of the equation ax — by = c.
401. Having given one set of integral values which
satisfy the equation ax + by = c, to find all other possible
integral solutions.
Let x = a, y = /8 be one integral solution of the
equation ax\by = c\ then aa + 6/8 = c. Hence, by sub
traction, a (a? — a) +6 (y — /8) = 0.
Now, since a divides a{x—aL), it must also divide
6 (y /8) ; a must therefore be a factor of y — /8, since it is
prime to 6.
Let then y—^^ma, where m is any integer; then
a (aj — a) = ~ 6 (y — /8) = — ma6 ; and therefore a? = a — mb.
Hence, if a? = a, y = /8 be one solution in integers of the
equation oa? — 6y = c, all other integral solutions are given
by
a? = a~m6, y = fi + ma,
where m is any integer.
INDETERMINATE EQUATIONS. 605
From the above, together with Art. 399, it follows that
there are an indefinite number of sets of integral values
which satisfy the equation ax\by = c. The number of
positive integral solutions of the equation is, however,
limited in number.
402. To find the nvmber of positive integral solutions
of the equation ax + by^c.
We have proved in Art. 399, that the equation
acc + by = c is satisfied by the values x = cq, y = — cp, or
by the values x = ^cq, y = cp, where p/q is the penultimate
convergent to a/6.
First suppose that x^cq, y = ^cp satisfy the equa
tion; then all other integral values which satisfy the
equation are given by
x=cq'mb, y=i — op + 7na (i),
where m is any integer.
From (i) it is clear that in order that x and y may both
be positive, and not zero, m must be a positive integer,
and that the greatest permissible value of m is l(^]
and its least value /{ — j + 1, so that the number of
diflferent values of m is I ('^) — I (—) . Hence, as one
set of values of x and y corresponds to each value of m,
the number of solutions is/(~] — /( — ) .
"ff'^if. +/./■■ =«■"» 'la) '" ^• ^ °'
/j — ^ — 1 according as^ is not or is less than^.
Thus the number of solutions is /fvj+1 or 1 1 A
506 IND£T£BMINAT£ EQUATIONS.
CO
according as the fractional part of ^ is or is not less than
the fractional part of — .
It can be shewn in a similar manner that if
w='cq,y = cp satisfy the equation, the number of solu
tions in positive integers is/frj + l or/f^J according
as the fractional part of — is or is not less than the frac
^ a
tional part of ^ .
Ex. 1. Find the positive integral yalnes of x and y which satisfy the
equation 7x  13t/ = 26.
7 111
We have 70 = 7 7 « 1 the penultimate convergent is therefore
i . Then 7 . 2  13 . 1=1; .% 7 (2 x 26)  13 (26)=26.
Hence one solution is a; =52, ^=26; the general solution is there
fore x=52 + 13m, y=26 + 7m.
[In this case the solution x=Oy t/=  2 can he seen hy inspection;
and hence the general solution is d;=13m, y=2 + 7m, which is
easily seen to agree with the previous result.]
Ex. 2. Find the positive integral values of x and y which satisfy the
equation Ix + lOy = 280.
7 111 2
Here ^ = t « « » the penultimate convergent being  , Then
7.310.2=1; /. 7(3.280) + 10(2.280)=280.
Hence x == 840, y =  560 is one solution in integers. The general
solution in integers is therefore a; =840 10m, y=i 560 + 7m; and,
in order that x and y may be positive m :^ 84 and m <: 80. Thus the
only values are a;=40, y=0; a;=30, y=7; a;=20, y=14; x=10,
2/=21; a;=0,y=28.
Ex. 3. Find the number of solutions in positive integers of the equation
3a; + 5y=1306.
3 111
Here = = 7 t 7;, whence 3.25.1=1;
5 1 + 1 + 2* *
.. 3 . (2 X 1306) + 5 (1306) = 1306.
Hence the general solution is a; = 2612  5m, y = 3m  1306.
For positive values of x and y we must have m>435 and m4>522.
Hence the number of solutions is 522  435=87.
J
INDETERMINATE EQUATIONS. 507
403. Integral sdutions of the two equations
ax + by + cz^dy a'x + b'y + c'z^d/
can be obtained as follows.
Eliminate one of the variables, z suppose; we then
have the equation
(a&  a'c)x + (b&  Vc) y^dc' d'c (i),
and this equation has integral solutions provided cu! — a^c
and be' — Vc are prime to one another, or will become
prime to one another after division by any common factor
which is also a factor of dc' — d'c.
Hence from (i) we obtain, as in the preceding articles,
the general solution
x = a + (be' — b'c) n, y = fi — (ac — a'c) n,
where a? = a, y = /8 is any integral solution, and n is any
integer.
Now substitute these values of x and y in either of
the original equations : we then obtain an equation of the
form Az + Bn = (7, from which we can obtain integral
solutions of the form z=^y+ Bm, n = S — Am, provided A
and B are prime to one another, or will become so after
division by any common factor which is also a factor of G.
Ex. Find integral solutions of the simultaneous equations
5x + 7y + 2z=24, 3a;y42=4.
Eliminating z, we have 13x+13y=52, or x+y=4. Whence
a;=2+n,y=2». Then6(2+n) + 7 (2n) + 2z=24,thatis«n=0.
Hence the general solution ia x=2knj y=2n, z=n,
11 Xf y and z are to he positive, the only solutions are x=4, y=0,
z=2; x=S,y=l,z=l; 8Jidx=:2,y=2,z=0; and, if zero values are
excluded, there is only one solution, namely x=B, y=lt z=l.
404. The following are examples of some other forms
of indeterminate equations. Other cases will be found in
Barlow's Theory of Numbers.
Ex. 1. Find the positive integral solutions (excluding zero values) of
the equation dx + 2y + 8z= 40.
It is clear that z cannot be greater than 4, if zero and negative
values of x and y are inadmissible.
608 INDETERMINATE EQUATIONS.
Henoe we have the following equations :
«=4, Sx + 2y= 8;
«=3, 8a; + 2y=16;
«=2, 3x + 2y=24;
2=1, 8a;+%=32.
And it will be found that all the solutions required are 2, 1, 4;
4, 2, 3; 2, 6, 3; 6, 3, 2; 4, 6, 2; 2, 9, 2; 10, 1, 1; 8, 4, 1; 6, 7, 1;
4, 10,1; and 2,13,1.
Ex. 2. Find the positiye integral solutions of the equation
6a;«13a^ + 6y«=16.
We have (3a; 2^) (2xSy) = 16; henoe, as x and y are integers,
3a;  2y must be an integer, and must therefore be a factor of 16.
Thus one or other of tiie following simultaneous equations must
hold good
3a;2y=±16, 2a;3y=± 1 (i) ;
3a;2y=± 8, 2a;3y=± 2 (ii) ;
3a;2y=± 4, 2a;3y=± 4 (iii);
3a;2t/=± 2, 2a;3y=± 8 (iv);
3a;2t/=± 1, 2a;3y=±16 (v).
Whence we find that 5x must be ±(482), ±(244), ±(128),
±(6 16) or ±(332).
Hence the only integral values of x are 4 and 2, the corresponding
values of y being 2 and 4.
Ex. 3. Solve in positive integers the equation
3a;> + 7a;y  2a;  6y  35 =0.
We have y (7x  5) + 3a;» 2a;  36=0;
3a;»2a;36 ^
•••y+ 7^^6— =^'
„ ^ a; 246 _
•••^^+^^+7F:r5=«5
.. 49y+21a;+l^^=0.
Hence ^ must be an integer, and therefore 7a;  5 must be a
7aj — o
factor of 1710. Whence it will be found that the only positive integral
solutions are x = 2, y = 3 and a; = l, y = 17.
EXAMPLES. 509
EXAMPLES XL.
1. Find all the positive integral solutions of the equations;
(1) 7a; +153^ = 59. (2) 8a; + 13^ = 138.
(3) 7a;+9y=100. (4) 15a; + 71y= 10653.
2. Find the number of positive integral solutions of
2a; + 3y = 133 and of 7a; + lly = 2312.
3. Find the general integral solutions of the equations
(1) 7a;13y=15. (2) 9a;lly = 4.
(3) 119a; 105^ = 217. (4) 49a; 692/= 100.
>
4. Find the positive integral solutions (excluding zero)
of the equations
(1) 2x+Sy + 7z=23. (2) 7a; + 42/+ I82; = 109.
(3) 5a; + y47« = 39, (4) 3a; + 2y + 3^? = 250,
2a; + 4y + 9« = 63. 9a;  4^^ + 5« = 170.
5. Solve in positive integers (excluding zero) the equa
tions :
(i) 2a^ 3a; + 2^=1329.
(ii) afxi/+2x3y = ll,
(iii) 2a;" + 5a!3/ 122/' = 28.
(iv) 2a;'  a^  y* + 2a; + 7y = 84.
6. Shew that integral values of a;, y and ^ which satisfy the
equation ax + by + cz = d, form three arithmetical progressions.
7. Divide 316 into two parts so that one part may be
divisible by 13 and the other by 11.
8. In how many ways can £1, 68, 6d, be paid with
halfcrowns and florins ?
9. In how many ways can JBIOO be made up of guineas and
crowns 1
510 EXAMPLES.
10. In how many ways can a man who has only 8 crown
pieces pay 1 1 shillings to another who has only florins?
11. Find the greatest and least sums of money which can
be paid in eight ways and no more with halfcrowns and florins,
both sorts of coins being used.
12. Find all the different sums of money which can be paid
in three ways and no more with fourpenny pieces and three
penny pieces, both sorts of coins being used.
13. Find all the numbers of two digits which are multiples
of the product of their digits.
14. Two numbers each of two digits, and which end with the
same digit, are such that when divided by 9 the quotient of each
is the remainder of the other. Find all the sets of numbers
which satisfy the conditions.
15. A man's age in 1887 was equal to the sum of the digits
in the year of his birth : how old was he 1
16. Shew that, if
then the number of solutions in positive integers (including
zero) of the equation a^x^ + a^^ + . . . + ajc^ = m, is A^y «i> <*8> • • •>
a^ being all integers.
The number of solutions of the equations x + 2y = n is
i{2w + 3 + (in.
At an entertainment the prices of admission were 1^., 2s. and
£5, and the total receipts £1000; shew that there are 1005201
ways in which the audience might have been made up.
17. The money paid for admission to a concert was £300,
the prices of admission being 5s., 38. and Is.; shew that the
number of ways in which the audieuce may have been made up
is 1201801.
CHAPTER XXX.
Probability.
405. The following is generally given as the defini
tions ot probability or chance: —
Definition. If an event can happen in a ways and fail
in b ways, and all these ways are equally likely to occur,
then the probability of its happening is t and the pro
bability of its failing is t .
To make the above definition complete it is necessary
to explain what is meant by * equally likely.' Events are
said to be eqvxilly likely when we have no reason to expect
any one rather than any other. For example, if we have
to draw a ball from a bag which we know contains
unknown numbers of black and white balls, and none of
any other colour, we have just as much reason to expect
a black ball as a white ; the drawing of a black ball and of
a white one are thus equally likely. Hence, as either a
black ball or a white ball must be chosen, the probability
of drawing either is J, for there are two equally likely
cases, in one of which tJie event happens and in the other
it fails. Again, if we have to draw a ball from a bag
which we Imow contains only black, white and red balls,
but in unknown proportions, we have just as much reason
to expect one colour as to expect either of the others, so
that the drawing of a black, of a white and of a red ball
512 PROBABILITY.
are all equally likely ; and hence the probability of draw
ing any particular colour is J, for there are three equally
likely cases, and any particular colour is drawn in one case
and is not drawn in the other two cases.
Another meaning may however be given to ' equally
likely ;' for events may be said to be equally likely when
they occur equally often, in the long run. For example, if
a coin be tossed up, we may know that in a very great
number of trials, although the number of ' heads ' is by no
means necessarily the same as the number of ' tails,* yet
the ratio of these numbers becomes more and more nearly
equal to unity as the number of trials is increased, and that
the ratio of the number of heads to the number of tails ¥^11
diflfer from unity by a very small fraction when the number
of trials is very great; and this is what is meant by saying
that heads and tails occur equally often in the long run.
Now, if each of the a ways in which an event can
happen and each of the h ways in which it can fail occur
equally often, in the long run, it follows that the event
happens, in the long run, a times and fails h times out of
every a + 6 cases. We may therefore say, consistently with
the former definition, that the probability of an event is the
ratio of the number of times in which the event occurs, in the
long nm, to the sum of the number of tiw£s in which events
of that description occur and in which they fail to occur.
Thus, if it be known that, in the long run, ont of eyery 41
children bom, there are 21 boys and 20 girls, the probability of any
21
particular birth being that of a boy is jr.
Again, if one of two players at any game win, in the long run,
5 games ont of every 8, the probability of his wiiming any particular
. 6
game IS ^.
We may remark that, in the great majority of cases,
including all the cases of practical utility, such as the data
used by Assurance Companies, the only way in which pro
bability can be estimated is by the last method, namely, by
finding the ratio of the actual number of times the event
PROBABILITY. 513
occurs, in a large number of cases, to the whole number
of times in which it occurs and in which it fails.
406. If an event is certain it will occur without fail
in every case : its probability is therefore unity.
It follows at once from the definition of probability
that if p be the probability that any event should occur,
1 — p will be the probability of its failing to occur.
When the probability of the happening of an event is
to the probability of its failure as a is to 6, the odds are
said to be a to 6 for the event, or 6 to a against it,
according as a is greater or less than h,
407. Exclusive events. Events are said to be
mutvjall/y exclusive when the supposition that any one
takes place is incompatible with the supposition that any
other takes place.
When different events are mutually exclusive the chance
that one or other of the different events occurs is the sum of
the chances of the separate events.
It will be sufficient to consider three events.
Let the respective probabilities of the three events,
expressed as fractions with the same denominator, be
5 S and ^
d' d ^"^ d'
Then, out of d equally likely ways, the three events
can happen in a^, a^ and a, ways respectively.
Hence, as the events never concur, one or other of
them will happen in a, + a, + a, out of d equally likely
ways. Hence the probability of one or other of the three
events happening is
?L±SjL^,thatis§ + ^» + §^.
cL a a a
This proves the proposition for three mutually ex
clusive events; and any other case can be proved in a
similar manner.
s. A. 33
514 PROBABILITY.
Ex. 1. Find the chance of throwing 3 with an ordinary sixfaced
die.
Since any one face is as likely to be exposed as any other face,
there is one favourable and five unfavourable cases which are all
equally likely; hence the required probability is ^ .
Ex. 2. Find the chance of throwing an odd number with an ordinaiy
die. Ans, .
Ex. 8. Find the chance of drawing a red ball from a bag which con
tains 5 white and 7 red balls.
Here any one ball is as likely to be drawn as any other; thus there
are 7 favourable and 5 unfavourable cases which are all equally
7
likely ; the required probability is therefore — .
Ex. 4. Two balls are to be drawn from a bag containing 5 red and 7
white balls ; find the chance that they wiU both be white.
Here any one pair of balls is as likely to be drawn as any other
pair. The total number of pairs is j2^2> &i^d the number of pairs
which are both white is ^G^: the required chance is therefore
7.6 / 12.11 7
1.2 / 1.2 ■'22*
Ex. 5. Shew that the odds are 7 to 3 against drawing 2 red balls
from a bag containing 3 red and 2 white balls.
Ex. 6. Three balls are to be drawn from a bag containing 2 black,
2 white and 2 red balls; shew that the odds are 3 to 2 against drawing
a ball of each colour, and 4 to 1 against drawing 2 white balls.
Ex. 7. A party of n persons take their seats at random at a round
table : shew that it is n  3 to 2 against two specified persons
sitting together.
408. Independent Events. The probability that two
independent events shovid both happen is the product of the
separate probabilities of their happening.
Suppose that the first event can happen in a^ and fail
in 6i equally likely ways; and suppose that the second
event can happen in a, and fail in b^ equally likely waya
Then each of the o^ + tj cases may be associated with each
of the Oy + b^ cases to make (oi + 6i) (oj, + tj) compound
cases which are all equally likely; and in a^a^ of these
compound cases both events happen. Hence the proba
PROBABILITY. 515
bility that both events happen is ; 7^7^ r\> that
is — ^j X — V , which proves the proposition.
Thus the probability of the concurrence of two inde
pendent events whose respective probabilities are p^ andp.
Cor. If pi and ^a be the probabilities of two inde
pendent events, the chance that they will both fail is
(1 — j9i)(l — Pa), the chance that the first happens and the
second fails is p^ (1 — jt),), and the chance that the second
happens and the first fails is (1 — p^p^*
It can be shewn in a similar manner that, if ©1, Pa,Pt»«"
be the probabilities of any number of independent events,
then the probability that they all happen will be p, .p, .p,. . .,
and that they all fail (1 Pi)(l — P2)(l — Ps)** &^c.
409. Dependent Events. If two events are not
independent, but the probability of the second is diflFerent
when the first happens from what it is when the first fails,
the reasoning of the previous article will still hold good
provided that p^ is the probability that the second event
happens when the first is known to have happened. Thus
if pi be the probability of any event, and p, the probability
of any other event on the supposition that the first has
happened; then the probability that both events will happen
in the order specified will be Pi^p^. And similarly for
any number of dependent events.
Ex. 1. Find the probability of throwing two heads with two throws of
a coin.
The probability of throwing heads is ^ for each throw; hence the
required probability is, by Art. 408, s ^ o = i •
Ex. 2. Find the probability of throwing one 6 at least in six throws
with a die.
33—2
516 PROBABILITY.
The probability of not throwing 6 is ? in each throw. Hence the
probability of not throwing a 6 in six throws is, by Art. 408, ( ^ ) t
and therefore the probability of throwing one six at least is
'©•
Ex. 8. Find the chance of drawing 2 white balls in succession from a
bag containing 5 red and 7 white balls, the balls drawn not being re
placed.
7
The chance of drawing a white ball the first time is j^ ; and,
haying drawn a white ball the first time, there will be 5 red and 6
white balls left, and therefore the chance of drawing a white ball
the second time will be jj . Hence, from Art. 409, the chance of
7 6 7
drawing two white balls in succession will ^ Tn ^ tt = na •
[Compare Ex. 4, Art. 407.]
Ex. 4. There are two bags, one of which contains 5 red and 7 white
balls and the other 3 red and 12 white balls, and a ball is to be
drawn from one or other of the two bags; find the chance of drawing
a red ball.
The chance of choosing the first bag is ^ , and if the first bag be
chosen the chance of drawing a red ball from it is ^h * hence the
1 f> i>
chance of drawing a red ball from the first bag is n ^ To = sr •
Similarly the chance of drawing a red ball from the second bag is
13 1
 X Y? — TTv . Hence, as these events are mutually exclusive, the
, . , . 6 1 37
chance required is ^ + — = ^ .
Ex. 5. In two bags there are to be put altogether 2 red and 10 white
balls, neither bi^ being empty. How must the balls be divided so as
to give to a person who draws one ball from either bag, (1) the least
ohtmce and (2) the greatest chance of drawing a red baU.
[The least chance is when one bag contains only one white ball,
and the greatest chance is when one bag contains only one red ball,
the chances being  and jt respectively.]
PROBABILITY. 517
410. When the probability of the happening of an
event in one trial is known, the probability of its happen
ing exactly once, twice, three times, &c. in n trials can be
at once written dowD.
For, if p be the probability of the happening of the
event, the probability of its failing is 1— 1) = }. Hence,
from Art. 408, the probability of its happening r times
and failing n — r times in amy specified order is )**g*"^
But the whole number of ways in which the event
could happen r times exactly in n trials is ^C^, and these
ways are all equally probable and are mutually exclusiva
Hence the probability of the event happening r times
exactly in n trials is nC^p'^**"'.
Thus, if (p + qY be expanded by the binomial theorem,
the successive terms will be the probability of the happen
iug of the event exactly n times, n — 1 times, n — 2 times,
&c. in n trials.
Cor. I. To find the most probable number of successes
and failures in n trials it is only necessary to find the
greatest term in the expansion of (p + q)*.
Cor. II. The probability of the event happening at
least r times in n trials is
p*+ n.p* ^q + \ o ^ />*^a' + . . . f 7—;'= — pq *
1.2 ^ * jr n~r ^
Ex. 1. Find the chance of throwing 10 with 4 dice.
The whole number of different throws is 6*, for any one of six
numbers can be exposed on each die; also the number of ways of
throwing 10 is the coeflBcient of x^^ in (a;+a;« +...+««)* for this co
efficient gives the number of ways in whidi 10 can be made up by the
addition of four of the numbers 1, 2, ..., 6, repetitions being allowed.
Now the coefficient of x^^ in («+«*+. ..+jb^*, that is in a?* (^— V,
is easily found to be 80. Hence the required chance is
80 _ 6
6.6.6.6~«1*
Ex. 2. Find the chance of throwing 8 with two dice. Ana, ~ .
518 PROBABILITY.
Ex. 8. Eind the chance of throwing 10 with two dice. Ans. j^ •
Ex. 4. Find the chance of throwing 15 with three dice. Aru, j^»
Ex. 5. A and B each throws a die; shew that it is 7 : 5 that A*b throw
is not greater than B*b.
Ex. 6. A and B each throw with two dice : find the chance that their
tiirows are equal. . 73
^ 648'
Ex. 7. A and B have equal chances of winning a single game at tennis :
find the chance of A winning the 'set' (1) when A has won 5
games and B has won 4, (2) when A has won 5 games and B has won
8, and (3) when A has won 4 games and B has won 2.
^'"•(l). (2). (8)g.
Ex. 8. A and B have equal chances of winning a single game; and A
wants 2 games and B wants 8 games to win a match: shew tnat it is
11 to 5 that A wins the match.
Ex. 9. A and B have equal chances of winning a single game; and A
wants n games and B wants n+1 games to win a match : shew that
the odds on ^ are l+i^^il?!^) to II44M .
2.4. 6. ..2n 2.4.6...2n
3
Ex. 10. A*B chance of winning a single game against B iszi find the
5
chance of his winning at least 2 games oat of 3.
^"' 125
2
Ex. 11. A^B chance of winning a single game against B is : find the
3
chance of his winning at least 3 games out of 5. , 192
Ex. 12. What is the chance of throwing at least 2 sixes in 6 throws
with a die? 12281
^'"' 46666
Ex. 13. A coin is tossed five times in succession : shew that it is an
even chance that three consecutive throws will be the same.
Ex. 14. Three men toss in succession for a prize which is to be given
to the first who gets 'heads*. Find their respective chances.
.421
* 7* 7' 7*
PROBABILITY. 519
411. The value of a given chance of obtaining a given
sum of money is called the expectation.
If r is the chance of obtaining a sum of money M,
then the expectation is if x— — 7 .
For if E be the expectation in one trial, E(a{h) will
be the expectation in a + b trials. But the chance being
Y , the sum M will, on the average, be won a times in
a + 6
every a + b trials; and hence the expectation in a + b
trials is Ma. Hence E(a + b) = Ma, therefore
E=Mx ''
a+y
Thus the expectation is the sum which may be won
multiplied by the chance of winning it.
Ex. 1. A bag oontains 5 white baUs and 7 black ones. Find the
expectation of a man who is allowed to draw a ball from the bag and
who is to receive one shilling if he draws a black ball, and a crown if
he draws a white one.
7
The ohanoe of drawing a black ball is r^ ; and therefore the
expectation from, drawing a black ball is Id. The chance of drawing
a white ball is r^ ; and therefore the expectation from drawing a
white baU is 28, Id. Hence, as these events are exclusive, the whole
expectation is 2«. 9d.
Ex. 2. A parse oontains 2 sovereigns, 8 halfcrowns and 7 shillings.
VThat should be paid for permission to draw (1) one coin and (2)
two coins? Ans. (1) 48. ^d. (2) 9«. Id.
Ex. 3. Two persons toss a shilling alternately on condition that the
first who gets 'heads' wins the shilling: find their expectations.
Arts, Sd., 4d.
Ex. 4. Two persons throw a die alternately, and the firdt who throws
6 is to receive 11 shillings : find their expectations.
Ans. 6«., 58^
620 PROBABILITY.
412. Inverse Probability. When it is known that
an event has happened and that it must have followed
from some one of a certain number of causes, the deter
mination of the probabilities of the different possible
causes is said to be a problem of inverse probability.
For example, it may be known that a blaok ball was drawn from
one or other of two bags, one of which was known to contain 2
black and 7 white balls and the other 5 black and 4 white balls ; and
it may be required to determine the probability that the ball was
drawn from the first bag.
Now, if we suppose a great number, 2Nt of drawings to be made,
there will in the long run be N from each bag. But in N drawings
2
from the first bag there are, on the average, ^ N which give a blaok
u
ball; and in N drawings from the second bag there are ^N which
2
give a black ball. Hence, in the long run, ^N out of a total of
N\^N blaok balls are due to drawings from the first bag ; thus
the probability that the ball was drawn from the first bag is
l^^(l^+H'*^*"l'
We now proceed to the general proposition : —
Let P,, P,,..., P^ be the prcbabilitiea of the eadstence ofn
causes, which are mutually exclusive and are such that a
certain event must have followed from one of them; and let
Pif Pi» '"fPn ^^ ^^ respective probabilities that when one
of the causes Pj, Pj, ..., P, eadsts it will be followed by the
event in question; then on any occasion when the event is
known to have occurred the probability of the rth cause is
Let a great number iV of trials be made ; then the
first cause will exist in JV.P, cases, and the event will
follow in If.P^.p^ cases. So also the second cause exists
and the event follows in N .P^,p^ cases ; and so on.
Hence the event is due to the rth cause in JV.P^.p,.
PROBABILITY. 521
cases out of a total of N'(Pj>^ + P^p^ + ... + Pj>J; the
p p
probability of the rth cause is therefore =r^ .
Having found the probability of the existence of each
of the diiferent causes, the probability that the event
would occur on a second trial can be at once found.
For let P/ be the probability of the existence of the
rth cause; then p^ is the probability that the event will
happen when the rth cause exists ; and therefore P/ .pri^
the probability that the event will happen from the rth
cause.
Hence, as the causes are mutually exclusive, the
probability that the event would happen on a second
trial is
Ex. 1. There are 8 bags which are known to contain 2 white and 3
black, 4 white and 1 black, and 8 white and 7 black bidls respectively.
A baU was drawn at random from one of the bags and found to be a
black baU. Find the chance that it was drawn from the bag con
taining the most black balls.
1 8 17
Here P^=Pj^=P^:=^, Also l>i=g, l>2=g and 1'8=Jq
1 7_
8 * 10 7
Hence the required probability is , ..... = t? •
131117 16
8* 6'*"3* 6'*"3' 10
Ex. 2. From a bag which is known to contain 4 balls each of which is
just as likely to be black as white, a ball is drawn at random and
foond to be white. Find the chance that the bag contained 3 white
and 1 black baUs.
The bag may have contained (1) 4 white, (2) 8 white and 1 black,
(8) 2 white and 2 black, (4) 1 white and 8 black, and (5) 4 black; and
the chances of these are respectively ^^ , ^^ , g , — and jg .
Art. 410. Also the chances of drawing a white btJl in these
3 11
different cases will be 1, j , ~ , ^ ^^^ respectively.
± 3
Ifi * i ^
Hence the required probability = j i =  .
134 1614o
i6'^i'16'*"2*l6'*"4'i6
622 PROBABILITT.
413. Probability of testimony. The method of
dealing with questions relating to the credibility of wit
nesses will be seen from the following examples :
Ex. 1. A ball has been drawn at random from a bag containing 99
black balls and 1 white ball; and a man whose statements are
aocorate 9 times oat of 10 asserts that the white ball was drawn*
Find the chance that the white ball was really drawn«
The probability that the white ball will really be drawn in any case
is jgg , and therefore the probability that the man will tnUy assert
1 9
that the white ball is drawn is Y7^^ x tk •
99
The probability that the white ball will not be drawn is ^r^ , and
therefore the probability that the man will faUely assert that the
99 1
white ball is drawn is =^ ^ in •
Hence as in Art. 412 the required probability is
JL A
100 ^ 10 1
'l 9 99""^^12'
100 ^ 10 "*" 100 ^ 10
Ex. 2. From a bag containing 100 tickets numbered 1, 2, ..., 100
respectively, a ticket has been drawn at random ; and a witness,
whose statements are accurate 9 times ont of 10, asserts that a
particular ticket has been drawn. Find the chance that this ticket
was really drawn.
In 1000 7<r trials the ticket in question will be drawn ION times;
and the witness will correctly assert that it has been drawn 9^ times.
The ticket will not be drawn in 990N cases, and the witness will
make a wrong assertion in 99^ of these cases ; but there are 99 ways
of making a wrong assertion and these may all be supposed to be
equally likely; hence the witness will wrongly assert that the
particular ticket has been drawn in N cases. Hence the required
9
probability is r? , so that the probability is in this case equal to the
probability of ^e witness speaking the truth.
Ex. 8. A speaks the truth three times out of four, and B five times
out of six ; and they agree in stating that a white ball has been drawn
from a bag which was known to contain 1 white and 9 bladk balls.
Find the diance that the white ball was really drawn.
The probability that the white ball will be drawn in any case ia
PROBABILITY. 523
Y^ , and therefore the probability that A and B will agree in truly
13 5
aflserting that a white ball is drawn ^ TH ^ 7 ^ r *
The probability that a black ball will really be drawn in any
9
case is j^; and therefore the probability that A and B will agree in
9 11
falsely asserting that a white ball is drawn ^ Ta ^ 7 ^ g •
Hence, as in Art. 412, the required probability is
13 5
10 ^ 4 ^ 6
5
T
10
3 6 9
^ 4 ^ 6 "*" 10 ^
T" 1
8'
Ex. 4. A speaks truth three times out of four, and B five times out of
six ; and they agree in stating that a white ball has been drawn from
a bag which was known to contain 10 balls all of different colours,
white being one. What is the chance that a white ball was really
drawn?
The probability that the white ball will really be drawn in any
case is ^ , and therefore the probability that A anii B will agree in
truly asserting that the white ball is drawn v&TK'^i'^a^Ta'
sM 4 o lo
The probability that the white ball will not be drawn in any case
9 1
is TTT . The probability that A will make a wrong statement is 7 ;
lu 4
hence, as there are nine ways of making a wrong statement which
may all be supposed to be equally likely, the chance that A will
wrongly assert that a white ball is drawn is 7 x q • Therefore the
chance that A and B will agree in falsely asserting that a white ball
is drawn is
9 111
T^ X , —  X
10 4xy 6x9 2160'
Ifi 1 *?I5
Hence the required probability is j = r^ .
1 1 I0O
16'*'2i60
Ex. 6. It la 8 to 1 that A speaks truth, 4 to 1 that B does and 6 to 1
that C does : find the probability that an event really took place
which A and B assert to have happened and wMch G denies ; the
event being, independently of tMs evidence, as likely to have
happened as not. Am, ].
624 PROBABILITY.
414. We shall conclude this chapter by considering
the following examples, referring the reader who wishes
for fiiller information on the subject of Probabilities to
the article in the Encyclopaedia Britannica, and to Tod
hunter's History of the Mathematical Theory of Proba
bility.
Ex. 1. A bag contains n balls, and all nnmberB of white balls from
to n are equally likely; find the chance that r white balls in succes
sion will be drawn, the balls not being replaced.
The chance that the bag contains s white balls is — Tt ^^^ the
chance that r balls in succession will be drawn from a bag contain
ing n balls of which « are white is —7 — =4—4 , . •
n(nl)...(nr+l)
Hence the chance required is
(n(nl)...(nr+l) , (n1) (n2)...(nr)
1 f n(nl)...(nr+l) (n1
w+1 l»(»l)...(nr+l) n(n
+ ...
l)...(nr+l)
^ r(rl)...l )
n{nl) ...\nr+l)\'
Now {1.2...r} + {2.3...(r+l)} + ... + {(nr+l)...(iil)n}
= (nr4l)(nr + 2)...n(n+l) ^^ 3^3
Hence the required chance is — ^ > ^^<}h is independent of the
whole number of balls in the bag.
If it be known that r white balls in succession have been drawn,
the probability of the next drawing giving a white baU can be at
once found from the preceding result.
For in a great number N^ of cases, there will be r white balls in
N N
succession in — = cases, and r+1 white balls in succession in — ^
r+1 r+2
cases. Hence the required chance is — ^ 7 — r = — jj .
r+^ r+1 r+J
Ex. 2. Two men A and B^ who have a and h counters respectively to
begin with, play a match consisting of separate games, none of which
can be drawn, and the winner of a game receives a counter from the
loser. Find their respecti'^re chances of winning the match, which is
supposed to be continued imtil one of the players has no more
counters, the odds being p : q l^at A wins any particular game.
PROBABILITY. 525
Let A'b chance of nltimate success when he has n counters be ti^»
Then A* a chance of winning the next game is ^ , and his chance
of nltimate success will then be Unrii* ^^^ ^*^ chance of losing the
next game is — ^ , and his chance of nltimate success will then be
p+q
Hence u^ss^— m. , + — ^ m„_, ;
•*• i'"ii+i~(P + ?)**« + 5^*111= ^» ^J^*^™ which it follows that m^
A + Bx
will be the coefficient of a;* in the expansion of ; r i, ,
p(p + q)x + qx^
provided A and B be properly chosen.
Now — . 7 r can be expressed in the form
p^(p + q)x + qx^ pqx 1x*
and hence the coefficient of a* is D + — (  ) .
P\pJ
Thos t*^=D +  fj , where C and D have to be determined.
But it is obvious that A*b chance of winning is zero if he has no
counters and unity if he has a+b, so that U0=O and Ua+b=^ ; hence
=!>+ — , and l=D + (  , whence the values of C and D
P P \PJ
are found, and we have
H'©i / {'(in
Hence A*b chance of winning the game is
i'(i)i/i'en
Similarly B's chance of winning the game is
{©'(/{'en
EXAMPLES XLI.
1. A and B throw alternately with two dice, and a prize
is to be won by the one who first throws 8. Find their
respective chances of winning if A throws first.
526 EXAMPLES.
2. A, jB and C throw alternately with three dice, and a
prize is to be won by the one who first throws 6. Find their
respective chances of winning if they throw in the order Ay
£,C.
3. Three white balls and five black are placed in a bag,
and three men draw a ball in succession (the balls drawn not
being replaced) until a white ball is drawn : shew that their
respective chances are as 27 : 18 : 11.
4. What is the most likely number of sixes in 50 throws
of a die 1
5. Shew that with two dice the chance of throwing more
than 7 is equal to the chance of throwing less than 7.
6. In a bag there are three tickets numbered 1, 2, 3.
A ticket is drawn at random and put back; and this is done
four times: shew that it is 41 to 40 that the sum of the
numbers drawn is even.
7. From a bag containing 100 tickets numbered 1, 2,
3,... 100, two tickets are drawn at random; shew that it is 50
to 49 that the sum of the numbers on the tickets will be odd.
8. There are n tickets in a bag numbered 1, 2, ..., n. A
man draws two tickets together at random, and is to receive a
number of shillings equal to the product of the numbers he
draws : find the value of his expectation.
9. An event is known to have happened n times in
n years : shew that the chance that it did not happen in a
particular year is ( 1 — ) .
10. If p things be distributed at random among p persons ;
shew that the chance that one at least of the persons will be
void is — \= .
11. A writes a letter to B and does not get an answer;
assuming that one letter in m is lost in passing through the
post, shew that the chance that B received the letter is
^ — — y , it being considered certain that B would have answered
the letter if he had received it.
EXAMPLES. 527
12. From a bag containing 3 sovereigns and 3 shillings,
four coins are drawn at random and placed in a purse; two
coins are then drawn out of the purse and found to be both
sovereigns. Shew that the value of the expectation of the
remaining coins in the purse is 11«. QcL
13. From a bag containing 4 sovereigns and 4 shillings,
four coins are drawn at random and placed in a purse; two
coins are then drawn out of the purse and found to be both
sovereigns. Shew that the probable value of the coins left in
the bag is 29^ shillings.
14. If three points are taken at random on a circle the
chance of their lying on the same semicircle is J.
15. A rod is broken at random into three pieces : find the
chance that no one of the pieces is greater than the sum of the
other two.
16. A rod is broken at random into four pieces : find the
chance that no one of the pieces is greater than the sum of the
other three.
17. Three of the sides of a regular polygon of 4w sides are
chosen at random; prove that the chance that they being
produced will form an acuteangled triangle which will contain
the polygon is ,. rfh kt •
^ ^^ (4w  1) (471  2)
18. Out of m persons who are sitting in a circle three are
selected at random; prove that the chance that no two of
those selected are sitting next one another is 7 tvt 7^ •
^ (ml)(m2)
19. If m odd integers and n even integers be written down
at random, shew that the chance that no two odd numbers are
In \n+ 1
adjacent to one another is , — =, m being 'i> n+1.
20. If m things are distributed amongst a men and b
women, shew that the chance that the number of things
received bv the group of men is odd, is ^r ^ ^ — .^i — ^ .
528 EXAMPLES.
21. The sum of two whole numbers is 100; find the chance
that their product is greater than 1000.
22. The sum of two positive quantities is given; prove
that it is an even chance that their product will not be less
than threefourths of their greatest product; prove also that
the chance of their product being less than onehalf their
1
greatest product is 1 — 7^.
23. Two men A and B have a and b counters respectively,
and they play a match consisting of separate games, none of
which can be drawn, and the winner of a game receives a
counter from the loser. The two players have an equal chance
of winning any single game, and the match is continued until
one of the players has no more counters. Shew that A*s chance of
winninfi^ the match is 7 .
24. An urn contains a number of balls which are known
to be either white or black, and all numbers are equally likely.
If the result of p + q drawings (the balls not being replaced) is
to give p white and q black balls, shew that the chance that the
0^ + 1
next drawing will give a black ball is — ^ ^^ .
° '^ p + q\2
25. Two sides play at a game in which the total number
of points that can be scored is 2m + 1 ; and the chances of any
point being scored by one side or the other are as 2m + 1 — aj
to 2m + 1 — y, where x and y are the points already scored by
the respective sides. Shew that the chance that the side
which scores the first point will just win the game is
(2m! 2m +1!)^
(m!)'m+ll 4m+ir
CHAPTER XXXL
Determinants.
415. If there are nine quantities arranged in a square
as under :
«!
a.
«.
6.
h
K
c.
c.
c.
then all the possible products of the quantities three to
gether, subject to the condition that of the three quantities
in each product one and only one is taken from each of
the rows and one and only one from each of the columns,
will be
a,V8» «A^8' ^aVi> ^A%> «8^i^a» ^^^ ^A^t'
Let now these products be considered to be positive or
negative according as there is an even or an odd number
of inversions of the natural order in the suflBxes ; then the
algebraic sum of all the products will be
«i^<58  ^A^2 + ^A^i  ^s^^s + %K<^2  %K^i (A.) ;
for there are no inversions in afi^c^, there is one inversion
in aj6gCj since 3 precedes 2, there are two inversions in
ajb^c^ since 2 and 3 both precede 1, there is one inversion
in ajb^c^ since 2 precedes 1, there are two inversions in
a,6jCj since 3 precedes both 1 and 2, and there are three
inversions in ajb^c^ since 3 precedes both 1 and 2 and 2
precedes 1.
s. A. 34
530
DETERMINANTS.
The expression (A) is called the determinant of the
nine quantities a^ a^, &c., which are called its elements;
and the products afi^c^, ^A^2» ^^* ^® called the terms of
the determinant.
416. Definition. If there are n* quantities arranged
in a square as under :
a. a.
a.
b, b. k
a.
m^ m, m.
m.
the members of the same row being distinguished by the
same letter, and the members of the same column by the
same sufl&x ; and if all the possible products of the quan
tities 71 at a time are taken subject to the condition that
of the n quantities in each product one and only one is
taken from every row and one and only one from every
column, and if the sign of each product is considered to be
positive or negative according as there is an even or an odd
number of inversions of the natural order in the suffixes;
then the algebraic sum of all the products so formed is
called the determinant of the n^ quantities or elements.
To denote that the n^ quantities are to be operated
upon in the manner above described, they are enclosed by
two lines, as in the above scheme.
The diagonal through the lefthand top comer is called
the principal diagonal; and the product of the n elements
ttj, 6,, Cg, ,m^ which lie along it, is caMed the principal
term of the determinant.
All the other terms can be formed in order from the
principal term by taking the letters in their alphabetical
order and permuting the suffixes in every possible way:
on this account a determinant is sometimes represented
by enclosing its principal term in brackets ; thus the
above determinant would be written [aj6,c,...mj, the
DETERMINANTS. 531
determinant is also often represented by the notation
When only one determinant is considered it is
generally denoted by the symbol A.
A determinant is said to be of the nth order when
there are n elements in each of its rows or columns, aod
therefore also n elements in each of its terms.
417. SiDce there are as many terms in a determinant
of the nth order as there are permutations of the w sufl&xes,
it follows that there are \n terms in a determinant of the
nth order. There are, for example, six terms in a deter
minant of the third order.
418. The law by which the sign of any term of a
determinant is found is equivalent to the following :
Take the elements in order from the successive rows
beginning at the first; then the sign of any term is positive
or negative according as there is an even or an odd number
of inversions in the order of the colwmns from which the
elements are taken.
We will now shew that the words row and column may
be interchanged in the above law. To prove this, consider
any product, for example, cbj>/)^d^e^^ and its equivalent
^i/2^8^A^6> where in the first form the letters follow the
alphabetical order and in the second form the numbers
follow the natural order.
We have to shew that the number of inversions
in the suffixes in the first form is the same as the number
of inversions of the alphabetical order in the second form.
This follows immediately from the fact that if, in the first
form, any suffix follow r suffixes greater than itself; then,
in the second form, the letter corresponding to that suffix
must precede r letters earlier than itself in alphabetical
order. Thus, in the example, 2 follows four suffixes greater
than itself in d^Ji^d.eJ^^, and /precedes four letters earlier
than itself in cjjj^dfl^e^.
34—2
532
DETERMINANTS.
Since the words rows and columns are interchangeable
in the law which determines the sign of any term, we have
the foUowing
Theorem. A determinant is unaltered by changing its
rows into columns and its columns into rows.
For example
a, ft,
a, b,
0^8 K
a, a.
a.
b. b. k
Ex. 1. Count the number of inversions in 2314, 3142 and 4231.
Jns, 2, 3y 5.
Ex. 2. Count the number of inversions in 4132, 35142 and 531264.
Ans. 4, 6, 7.
Ex. 8. What are the Bigns of the terms b/g, cdh and ceg in the
determinant a b c
d e f
g h k
[The order of the columns is 231, 312 and 321.]
Aim. +, +, — .
Ex. 4. What are the signs of the terms bgiq, celn and dfkm in the
determinant abed
e f g h
i 3 h I
m n p q
[The order of the columns is 2314, 3142 and 4231.]
Atis, +, — , — .
419. Theorem I. If in any term of a determinant any
two suffixes be interchanged, another term of the determinant
will be obtained whose sign is opposite to that of the original
term.
Let P .ha.kfi be any term of a determinant, P being
the product of all the elements except ha and k^ ; then, by
interchanging a and fi we have P.hp.ka Now since
P .ha.kp is a term of the determinant, P can contain no
element from the rows of h*s and k's and no element from
DETERMINANTS. 533
the a or ^ columns ; and this is a sufficient condition that
Phfika should also be a term of the determinant.
We have now to shew that the two terms have
different signs.
First suppose that two consecutive suffixes are inter
changed.
Consider the term AhJcpB where A denotes the product
of all the elements which precede ha and B the product of
all the elements which follow kp. By interchanging a and
13 we have AhpkaB, which we have already found is a term
of the determinant.
Now the number of inversions in the two terms must
be the same so far as the suffixes contained in A, or in B,
are concerned, whether compared with one another or with
a and ^ ; but there must be an inversion in one or other of
a^ and ^a but not in both. Hence the numbers of
the inversions in the two terms differ by unity, and therefore
the signs of the terms must be different.
Now suppose that two nonconsecutive suffixes are
interchanged ; and let there be r elements between the two
whose suffixes, a and /3 suppose, are to be interchanged.
Then a will be brought into the place of ;8 by r+1 in
terchanges of consecutive suffixes, and /3 can then be brought
into the original place occupied by a by r interchanges
of consecutive suffixes ; and therefore the interchange of
a and )8 can be made by means of 2r + 1, that is by an odd
number, of interchanges of successive suffixes. But, by the
first case, each such interchange gives rise to a loss or gain
of one inversion ; and hence there must on the whole be a
loss or gain of an odd number of inversions : the sign of the
new term will therefore be different from the sign of the
original term.
420. Theorem II. A determinant is unaltered in
absolute voLtie, but is changed in sign, by the interchange
of any two colwrmis or any two rows.
Suppose that in any determinant the rows in which
the letters h and k occur are interchanged. Then, if
534
DETERMINANTS.
A ,ha.B.kp,G he any term of the original determinant,
the term of the new determinant formed by the elements
which occur in the same places as before will be AkaBhfiG;
and these two terms must have the same sign in the two de
terminants. Now by Art. 419 we know that A.ka.B.hp.G
is a term of the original determinant and that its sign
is different from that of A ,ha,B,kfi.G. Hence any
term of the new determinant is also a term of the original
determinant but the sign of the term is different : the two
determinants must therefore be equal in absolute magni
tude but different in sign.
The proposition being true for rows is, from Art 418,
true also for columns.
For example
«! flt <^
= —
h 6a 6,
Ci Ca c^
Oi a, Oa
=
«i «• «a
h 6, ^1
Ci </g Cj
«1 Cs Cj
6i b^ 6j
421. Theorem III. A determinarU, in which two
rows or two cohmins a/re identical, is equal to zero.
When two rows (or two columns) are identical, the
determinant is unaltered either in sign or magnitude by
the interchange of these two rows (or columns). But, by
Theorem II, the interchange of any two rows (or columns)
of a determinant changes its sign. Thus the determinant
is not altered in value by changing its sign: its value
must therefore be zero.
Ex. 1. Find the value of
1
a a'
1
6 6«
1
c c*
It is obyioos that two rows would become identical, and therefore
the determinant would vanish, if a =6. Hence A must be equal to
an expression which has a  6 as a factor. Similarly hc and <; — a
must be factors of A. But A is by inspection seen to be of the third
degree in a, &, c ; hence AsX(&c)(ca)(a6), where L is
numerical. The principal term of A is 5c' and this is the only
term which gives be\ and the coeflBcient of 6c' in i (6  c) (c  a) (a  h)
isL; therefore L=l. Thus A=(6c) (co) (a 6).
DETERMINANTa
535
Ex. 2. Find the value of
1
1
1
1
a
h
c
d
2
a
Ans. (bc){ca)(ah)(ad){bd){cd).
Ex. 3. Find the value of
1
1
1
1
a
b
e
d
d2
a*
iifw.  (bc) (c  a) (ab) {a d) {b  d) (c  d) (a+b + c + d).
422. Theorem IV. //*aS the elements of one row or
of one column of a determinant be multiplied by the same
qm/atity^ the whole determinant wiU be multiplied by that
quantity.
For every term of the determinant contains one
element and only one from each column and from each
row ; and it therefore follows that if all the terms of one
row or of one column be multiplied by the same quantity,
every term of the determinant, and therefore the sum of
all the terms, will be multiplied by that quantity.
Cor. From the above, together with Theorem III, it
follows that if two rows or two columns of a determinant
only differ by a constant factor, the determinant must
vanish.
For example
mbi fnc.
V<h pbf
Also
PCt
ma
nib
=wi7ip
na
Tib
mc nc
1
1
1
h Cj
a
b
c
a
b
c
1
1
1
moi
IlkZjl
=0.
nbi
n5«
pci
423. Minor determinants. When any number of
columns and the same number of rows of a determinant
are suppressed, the determinant formed by the remaining
elements is called a minor determinant
536
DETERMINANTS.
A miDor determinant is said to be of the first order, or
to be a first minor , when one column and one row are
suppressed ; it is said to be of the second order, or to be a
secmd minor, when two columns and two rows are sup
pressed ; and so on.
The determinant obtained by suppressing the line and
the column through any particular element is called the
minor of that element, and will be denoted by A^, where x
is the element in question.
Thus
a, Oj
«3 ^3
(h
at,
6,
and
h
are first minors of
and are A^ , Aj and A^ respectively.
424. Development of determinants. Consider the
determinant of the fourth order
A =
^1 ^2 ^8 ^4
6. 6„ 6„ k
d, d^ d„ d.
A certain number of the terms of A will contain a
1 >
let the sum of all these terms be a^,A^, Similarly let
the sum of all the terms which contain a,, a, and a^, be
respectively a^ . A^, a^ . A^ and a^ . A^, Then, since no term
can contain more than one of the letters a^, a^, a,, a^ we
have
A = a^A^\ a^^ + asA^\ a^A^ (i).
Now, since no term of A which contains a^ can contain
any element from the column or the row through Oj, it
follows that every term of A which contains Oj is the
product of Oj and some term of A^^; conversely the product
of «! and any term, T, of A^^ will be a term of A, and the
sign of the term a^,T of A will be the same as the sign of
the term T of A^^, for there is no change in the number of
DETERMINANTS.
537
inversions. Hence the sum of all the terms of A which
contain a^ is Oj. A^.
So also, every term of A which contains a^ is the
product of 0, and some term of A^,, and the product of a^
and any term, T, of A^, will be a term of A, but there is
one more inversion in the term Oj . jT of A than there is in
the term T of A^, since 2 precedes 1. Hence the sum of
all the terms in A which contain o^ is ■ o^ . A^,.
Similarly the sum of all the terms of A which contain
a, are a^ . A^; and the sum of all the terms which con
tain a^ are — a^ . A^^.
Hence
A = ai.A«,a^.A^ + as.Aa.a4.Aa, (ii).
By means of Articles 419 and 420, we can shew in a
similar manner that
A =  6i Aft, + 62 Aft,  6, Aft. + 6, Aft,
= aiAa,6iAft^ + CiAe,(iiAd, = &c.
Cor. By comparing (i) and (ii) we see that the co
factors of the elements a^, a,, &c., are equal in absolute
magnitude to the minors of the same elements.
425. We have in the previous article considered the
case of a determinant of the fourth order; the reasoning is
however perfectly general, so that if A be a determinant
of the nth order having a^, a,,..., a^ for the elements of its
first row or column; then will
A = ai.A«^aaA^H... + (l)"^a„A«..
So also
A = (irMA^i.A*,Ar,.A;fc. + ...H(ir^Ar„AJ.
Where i^, A?,,..., k^ are the elements of the rth row.
For example
6,
«s
= 0,
h h
h
<^9 Cs
^8
%! ^]
+ a.
h ftj
=^1 (^2^8  fejC^  Oa (61C3  Vi) + «8 (Va  Vi)»
538
DETERMINANTS.
Prove the foUowing :
X.
1
2
1
1
3
2
1
1
2
2
1
3
2
1
1
8
2
1
=4.
= 18.
2 =27.
2
1
= 16.
7.
a c
=2abc.
a b
b c
a+b e c
a b+c a
b b
c + a
6.
= iabc.
1 2
2 1
2 2
6 5 5
5 6 5
5 5 6
a, a a,
abb
a b c
b + c e b
e c+a a
b a a+b
= a (6  c) (a  b).
=4a&c.
3 2 2 2
2 3 2 2
2 2 8 2
2 2 2 3
= 9.
lO.
1 1
1 2
1 3
1 4
1 1
3 4
6 10
10 20
= 1.
11. Write down the cofactors of a, f and c in the expansion of
the determinant A= a h g
h b f
9 f c
Shew that, if A^ B, &o, are the oofactors of a, 6, &o, in the
above determinanti and A\ B', <fec., the cofactors of A, B, &o, in
the determinant
A H
G
H B F
F O
*!, ^' ^'
; then — =s — =
a b
=A.
426. Theorem V. If the elements of one column of
a determinant be multiplied in order by the cofactors of
the corresponding elements of any other column ; then the
sum of the products will be zero.
Let the elements of the rth column be multiplied by
the cofactors of the corresponding elements in the «th
column ; then the sum of the products will be
a^.A, + b^,B^+.,,
DETERMINANTS.
589
Now consider the determinant which differs from the
original determinant only in having the sth column
identical with the rth; then A„ B„ &c. will be the same
in the new determinant as in the original one.
The value of the new determinant will therefore, by
Art 425, be equal to
since a^ = a„ b^ = 6„ &c.
But, from Art 421, we know that the new determinant
is zero.
Hence a^.4, + 6^.jB,+ ... =0.
Thus in the determinant ^=[(iih^^4]i we have
A= OiAi + a^A^ + Oj^a + ^444 ;
also 0= biAi + h^d^ + b^A^ + 64ii4 ,
= Oj^, + 6iBg + CiO, + djDg , Ac.
427. Theorem VI. If each element of any row (or
colwmn) of a determinant he the sv/m of two quantities^ the
determinant cam, be expressed as the sum of two deter
minants of the sa/me order.
It will be sufficient to take as an example the deter
minant
«! + «, K ^,
«• + ««
By Art. 424, we have, if A^, A^, A^, be the cofactors of
the elements of the first column,
(h h
+
^2
2.
c.
640
DETERMINANTS.
Similarly it can be proved that
Oi+oi 6,
A ci
=
«1
h Ci
+
«2"*«2 ^a^2 «a
ttj 63 Cj
Oj + ^S ^s^3 c$
Os ^8 <^S

<h A ci
—
«! A <Ji
<h ^ c^
Oa ^2 <^a
H A
Cz
03 A
c»
03
^3
428. Theorem VII. A determinant is not altered m
valvs by adding to all the elements of any column {or row)
the same multiples of the corresponding elements of any
number of other columns {or rows).
Take as an example a determinant of the third order :
we have to shew that
a,
a„
% + '^K + WC3 63 C3
By Theorem VI, the last determinant is equal to
^1
a.
c
c
+
m6j 6^ Cj
m6„ 6« c,
m6„ 6.
8
+
Ci
'8
But each of the last two determinants is zero [Art. 422,
Cor.] ; this proves the theorem.
Ex. 1. Shew that
=0.
1 a b + c
1 b c + a
1 c a+6
Add the second column to the thirds; then
A=
1 a
a + 6 + c
^(a^b + c)
1 a I
1 b a + b{c
16 1
1 c a + b\c
1 c 1
since two columns are now identical.
Ex. 2. Shew that
abed
a bed
a bed
= 8abcd»
rt b
c d
=0,
DETERMINANTS.
541
Add the first row to each of the others ; then
=:A
a b
c d
= a
26
2c 2d
= 2ab
2c 2d
26
2c 2d
2c 2d
2d
2c. 2d
2d
2d
= Sahcd,
Ex. 3. Shew that
= 0.
a + 26 a+46 a+66
a + 36 a + Bb a+76
a + 4& a+66 a+86
Take the second row from the third, and then the first from
the second.
Ex. 4. Shew that
b + c
a
tc
a—b
= 8a6c.
bc c+a
ba
Ex. 5. Find the ya
cb
lue of
ca a+b
1 16 14
4
•
*
12 6 7 9
8 10 11 6
A=
1 16 14
13 i
4
i 2
=48
16
1
15
14
4
12 6 7 9
12 6
. 7
9
4 4 44
1 1
1
1
12 12 12 12
= 48 1 1
16 1
1 :
4 4
I 1
=0.
1
12
6 7 9
1
111
Ex. 6. Find the values of
111
and
3 2
2 2
•
4 5 11
2 3
2 2
3 9 4 1
2 2
3 2
— /
i 4
4
1
2
2
2 3
^n«. 0, 9.
429. The following is an impoitant example.
To shew that
«1
h
Cl
2
m
n
rr
«!
6i Ci
•
a. ft 7i
Oa
h
Ca
P
3
r
Oa
62 Cg
02 ft 72
«3
h
H
8
t
u
ag
6g C3
^3 ft 73
Ol
ft
7j
02
ft
7a
Os
ft
7t
542
DETERMINANTS.
It is in the first place dear that a term of the deteiminant of the
sixth order will be obtained by taking any term of [cLfi^ with any
term of [o^hyH' ^^^B A = [Oi^jcj . [04^71] together with tenns
inyolving 2, m, n, &o, ; and we have to shew liiat all terms involving
any of &e letters Z, m, n, <fec. will vanish.
Now, in eveiy term of the minor of 2, three elements most be
chosen from the last three rows, and two only of these can be chosen
from the last two columns ; hence one of the three elements mnst be
zero, and therefore every term of A is zero. Hence the minor of I,
and so iJso the minor of each of the elements m, n, <fec. is zero ; this
proves that there are no terms involving any of the letters Z, m, n, &o.
It can be proved in a similar manner that any determinant of the
2nth order is the product of two determinants of the nth order,
provided eveiy element of one of the nth minors of the original
determinant is zero.
430. Multiplication of determinants. We shall
consider the case of two determinants of the third order :
the method is however perfectly general.
To express as a determincmt of the third order, the
product of the two determinants.
A.=
a, 61 Ci
a.
and
A.=
«8 K 0,
We know from Art. 429 that
ai
13.
71
78
A,A,=
(h hi
Ci
1
1
A
1
7i
7,
7s
.(A).
Multiply the first three rows by o^, /S^, y^ and add the
products to the fourth row ; then multiply the first three
rows by a^, ^^, y^ respectively and add the products to the
fifth row; and then multiply the first three rows by a,, fi^y^
respectively and add the products to the sixth row. We
shall then have the equivjdent determinant
DETERMINANTS.
543
fli » &i , Ci , 1, ,
Oa » ^9 > ^2 , , 1,
flj . h » ^'s ,0,0,1
Oiai + aa/Si + OjYi, ftiOi + dj^i + ^sYi* ^lOi + ^A+Wi* ^ » <) » ^
aia,+aA+«s7t» ^«j+^+^7t» Cia,+Cj/3,+c,7„ , , '
«ia,+aaft+a,7„ 6i08+6A + ^87t» Ci08+C2/3,+c,7,, , ,
which is by Art. 429 equivalent to the product of
— 1 , that is 1, and
110
01
OiOi+Oaft+ajYi, bio^ + b^ + b^i, ^lOi+cA + CsVi
aia,+aA+a,7j, 6103+ 6^2+ 6,72 > CiO^+c^i+Csyi
aia,+a2/8,+a,7„ 6ia,+6a/3j+6878» Cia,+C2/%+C87,
Hence the required product is the determinant last
written.
Ex. 1. Multiply
X y z
by
a e h
z X y
h a c
y z X
e b a
The required product is
X Y Z
Z X Y
1
Y Z X
where X^ax+by+cz^
Y=ay\bz+ex^ and Z=az + bx+ey,
Since
ss «* + y ' + iB*  ^xyz^ and the other deteiminants
X y z
z X y
y z X
are of the same form, we see that the product of any two expressions
of the form a^+y^+i^3xyz oan be expressed in the same form.
[See Art. 156, Ex. 4.]
Ex. 8. Shew that
26c a«, c*,
c9, 2ac  6^,
b^,
62
a"
a^
Form the product of
a.
Ex. 3. Shew that
A,
, 6.
c, a,
6, c.
b
a
02
2a6c2
and
(a8+6a+c»3a6c)a
a.
c.
&i Ci
6 c
a &
6, c a
*, where ii, Pj, <fto.
a
8
are the cofactors of Oj , 6^ , &o. in the expansion of the determinant
(a,6acj.
544
DETERMINANTS.
For
^1
4s
B,
^2
Oj a.
[oiftac J
[oiVa]
since
and
Hence
^i6i + il2&j+il86,=&c. = [Art. 426],
431.
The notation
a.
a.
a. a.
&. h h K
is employed to denote the system of four determinants
obtained by omitting any one of the columns.
432. We conclude with the following important appli
cations of determinants.
Simultaneous Equations of the First degree.
The solution of any number of simultaneous equations of
the first degree can be at once obtained by means of the
foregoing properties of determinants.
First take the case of the three equations
Multiply the equations in order by A^, A^, A^, where
A , A , ilj are the cofactors of a,, a,, a^ respectively in the
determinant a^ b^ c^
a, K ^2
«8 ^8 ^8 .
Then we have by addition
(a,^, + a^A, + a,A,) x + {b,A^ + b^A^ + 63^3) y
+ (P,A^ + c^, + 03^3) z = k^A, + M2 + K^n ;
B£T£RMINANTS.
545
that is [a, 6, c^] x = [*, 6, c,],
for from Art. 426 the coefficients of y and z are zero.
Similarly we obtain
and [a^ ^a cj ^ = [a^ 6, k^.
Now consider n equations of the form
a^x^ + b^x^ + cjs^ + d^x^ + = i^.
As before, multiply the equations in order by J.^, A^,
A^y &c. the cofactors respectively of a^, a^, a^, &c. in the
determinant [aj 6^ c,. . .] ; then we have by addition
(a^A^ + a^A^ + ttj^, +. . .) a? = k^A^ + A?^,, + k^A^ +. . .,
the coefficients of y, z, &c. being all zero by Art. 426.
Hence ^ = [*iA^.
So also y = tvV^.j;] &c.
Ex. 1. Solve the equations
a; + 2y + 3z = 6,
2a;+4y+xf =7,
8a; + 2^ + 9^ = 14.
The values of a;, y, c are respectively
6 2 3
16 8
12 6
7 4 1
2 7 1
2 4 7
14 2 9
3 14 9
and
3 2 14
12 3
1 "
12 3
12 3
2 4 1
2 4 1
2 4 1
8 2 9
3 2 9
3 2 9
and it will be found that each determinant is  20, so that
Ex. 2. Solve the equations
X +y •\z +w +Aj =0,
flwj +6y +c« +dt(; +fc'=0,
o»a; + 6«y+c««+<i2u? + /w«=rO,
o»«+ d'y +c*« + d»w + A:*=0.
8. A. S5
546
DETBRMINANTS.
We have
x =
1
b
62
1 k
d k^
53 c' d» fc*
a
a'
a?
Ill
bed
62 c* d^
68 c* d3
ifc(cd)(d6)(6c)(fe6)(fec)(ferf) .
"" (cd)(d6)(6c)(a6)(ac)(ad) '
fc(fc6)(fec)(fed)
•'• *"" (a6)(ac)(ad) '
and the values of y, z and «; can be written down from that of x.
433. Elimination. To find the condition tluit the
three equations
a^x + 6,2/ + Cj = 0,
^2^ + ^2^ + Cg = 0,
agfl? + 632/ + ^a = ^>
may be simultaneously true.
Multiply the equations in order by (7^, (7,, O,, the
cofactors of c^, c,, Cj respectively in the determinant
Then by addition we have
a.
^
c,
a.
6.
c.
«.
t.
c.
(a,C, + aP^ + a3(73) ^ + (6,0, 4 6,0, + 63(73) y
+ c,(7, + c,(7, + cfi^ = 0,
that is, from Art. 426,
= 0,
a, 6, c,
a, 6, c,
«8 ^8 ^8
which is the required condition.
The three homogeneous equations Ojaj + 6,y+Ci;e;= 0305+ 632/ + C2«
= a^ + 68j^ + Cg« = are obviously satisfied by tne values a; = y=z=0.
If however or, y, 2; are not aU zero, it follows from the above that the
condition \a^^^=^ must hold good.
DETERMINANTS.
647
It can be shewn in a similar manner that the condition
that n equations of the form UjX + 6,y + ... + Aj = 0, with
(n — 1) unknown quantities, may be simultaneously true is
434 Sylvester's method of Elimination. This is
a method by which a? can be eliminated from any two
rational and integral equations in w. The method will be
understood from the following examples.
Ex. 1. Eliminate x from the equations
ax^+bx+c^O and j)a:'+ga; + r=0.
From the giyen equations we have
ax^+bsfi+cx =0,
aa?+bx+c=0,
px^+qx^\rx =0,
and px^+qx+r=0.
Now we may oonsider the dififerent powers of x as so many different
unknown quantities; and the result of eliminating x', x^ and x from
the four last equations is by Art. 433
a 6 <; =0.
a h c
p q r
p q r
[This result is equivalent to that obtained in Art 153, Ex. 3.]
Ex. 2. Eliminate x from the equations ax^+biifi'\cx+d=0 and
px^+qx+r=iO.
From the given equations we have
nxi^+bx^+cx^^dx =0,
aic^+biifi+cx + d=0,
pxt^+qx^ + rx* =0,
px^\qx^+rx =0,
2>a;* + ga5+r=0.
Eliminating x*, oc*, a?, x from the five last equations as if the
different powers of x were so many different unknown quantities, we
have the condition
a
P
3
P
c
b
r
P
d
c
d
r
= 0.
35—2
548
EXAMPLES.
EXAMPLES XLH.
X. Shew that
2. Shew that
3. Shew that
4. Shew that
6. Shew that
6. Shew that
7.' Shew that
Shew that
0. Shew that
lO. Shew that
ab
ca
1 a
1 h
1 c
h+c
ab
c^ + a^
cb
a^be
b^ca
c^db
c\a
he
a^+b^
=0.
=^a%h'^.
6'+c' c' + a'
b" + c
ft
c" + a"
=2
a + b\2c
c
e
a+h
a' + b'
a" + b"
a b
6+c + 2a 5
a c+a+26
b
o" b" c
= 2(a+6 + c)3.
a
a'
.11
c
c'
a^
a^^ab
ba
(6+c)«
6«
6c
ca + a^
be
b^
b^+bc
c»
{c+a)^
a«
bc + h^
ca
ah+a^ ab + b^
(a+6)a
ac + c^
=4a^h^cK
ca
be
(6+c)2
b^
a
b
c
a
c
b
b
c
a
ca
a5
a^
{c + ay
c«
c
b
a
ac
c»
6»
a^
(a + 6)«
ca+c^
a6
6c
a&
(c+a)»
a2
63
(a +6)2
1
1
1
1
ca
6»
= 2(6c + ca+a6)'.
= (6c + ca + a6)5.
= 2a6c(a+6 + c)5.
=2a6c(a + 6 + c)*.
1 1
c^ 6«
o^
a*
= (a+6+c)(a+6 + c)(6+c+a)(c + a+6),
11. Prove that
a^
6*
c«
a*
i8»
6« c«
aa
6)3
C7
aa
C7
6/3
6/3 C7
C7 6/3
aa
aa
EXAMPLES.
549
12. Shew that
18.
14.
1 1
1 1 + a
1 1
1 1
Shew that
1 + a 1 1
1 1 + 6 1
1 1 1+c
111
Shew that a b c
bad
^ d a
deb
1
1
1 + 6
1
1
1
1
1 + d
d
c
6
a
1
1
1
1+c
= abc.
^ ^ /, 1 1 1 1\
= a6cd(l +  +  +  + ).
\ a c dj
= (a + 6 + c + d)(a + 6cd)(a + c6d)(a+d6c).
15. Shew that
16. Shew that
1+a;
1
1
1
a
6
c
2
2+x
2 .
2
6
a
d
3
3
S + x
3
4
4
4
4 + aj
=a;S(a; + 10).
c
a
6
d
c
6
a
= (a2 + 62 + c2 + d2)3.
17. Shew that
1
1
1
1
a
6
c
d
o2 a5+6cd
62 b^ + cda
c* c'+iia6
d^ d' + a6<j
0.
18. Shew that
19. Shew that
20. Shew that
a a a
a
=a{b''a)\
aba
a
a a b
a
a a a
6
abb
6
= (a6)*.
aba
a
b b a
6
a a a
6
axby
cz
ay + bx
cx + az
ay +bx
byczax
bzkcy
cx+az
bzkcy
czaxby
= (a2 + 62 + c2) (a;«+2^«+2^ (axiby + cz).
650
EXAMPLES.
ai. Shew that
a« a2(6c)« 6c
= (6c)(ca)(a&)(a + 6+c)(a2 + 6« + c2).
c» c»(a6)»
. Shew that
(6c)a (a6)« (ac)«
(5 a)* (ca)^ (bc)*
{caf (c6)» (a6)a
= 2(a»+6« + c«6ccaa6)«.
28. Shew that, if any determinant yanisbes, the minors of any one
row will be proportional to the minors of any other row.
24. Shew that
a»+l
ba
ca
da
db
62 + 1
cb
db
ac
be
c«+l
de
ad
bd
cd
fP+1
= a2 + 6« + c« + d» + l.
25.
Shew that
1
1 as + a*
1 ab+afi
1 ac + ay
26.
o6 + oj8 ac + ay
6» + )^ 6c + i37
bc+py c'^ + y
Shew that the determinants
= (67  Cj8+ ca  07 + a/3  6a)*.
X
X
y
z
z
a
a
a
6
6
6
c
c
c
are all zero.
27. Shew that
28. Shew that
X c 6
c \ a
6 a \
29. Shew that
X y
a 6
d c
w z
x^yz
z^xy
v^zx
y^zx
x^yz
z'^xy
z^xy
y^zx
x^yz
X
y
z
y
z
X
z
X
y
a' + X* a6+Xc acX6
06 Xc 62 + X* 6c + Xa
ac + X6 6c Xa c*+X*
=XMX» + tt* + 6a + c»)».
z w
=
x+w y+z
»
xw yz
c d
a+d b+e
ad bc
6 a
y X
i
CHAPTER XXXII.
Theory of Equations.
435. Any algebraical expression which contains a? is
called a function of oo, and is denoted for brevity by / (x),
F{x), <f> (x), or some similar symbol.
The most geneml rational and integral expression
[Art. 75] of the nth degree in x may be written
a^'^ + a^x'^^ + a^^ + ... + a^,
where a^, Oi, aa,... do not contain x.
Since all the terms of any equation can be transposed
to one side, every equation of the nth degree in x can be
written in the form
where n is any integer, and the coefficients a©, ch, (h^
do not contain x.
Now any equation in a? is equivalent to that obtained
by dividing every one of its terms by any quantity which
does not contain x ; and, if we divide the left side of the
above equation by a©, the coefficient of a?^, we shall obtain
the equation of the nth degree in its simplest form,
namely
x"" +p^af^^ +)2«"2 + .•. +i)n = 0,
where pi, p^* Pz^* ^o not contain x, but are otherwise
unrestricted,
552 THEORY OF EQUATIONS.
436. If we assume the fundamental theorem* that
every equation has a root real or imaginary, it is easy to
prove that an equation of the nth degree has n roots.
For suppose the equation to be /(a:) = 0, where
fix) = of" +)iaJ^i ^p^^^ +...+;)«.
Since /(a?)= has a root, Oj suppose, we have /(Oi) =0,
and therefore [Art. 88] /{x) must be divisible by a? — Oj,
so that f{(io):^{X'a^^{aj\ where ^ (a?) is an integral
function of x and of the {n — l)th degree. Similarly,
since the equation <l>(x) = has a root, a^ suppose, we
have <l> (x) =^{x''a^'^ (x), where ^ (x) is an integral
function of x of the (n — 2)th degree. Hence
Proceeding in this way we shall find n factors o{ /(x) of
the form a? — Oi, and we have finally
f(x) = (a? — Oi) (a? — Oa). . .(x — a„).
It is now clear that Oi, a^,,.,, an are roots of the
equation f(x) = ; also no other value of x will make
f(x) vanish, so that the equation can only have these
n roots.
In the above the quantities 01,03,08,... need not be
all different from one another; but if the factors x a^,
a? — O2, a? — Os, &c. be repeated p, 5, r, &c. times respectively
in f(x), we must have
f(x) = (a;  Oi)^ (x  Oa)^ (x  asY. . .,
where p + q + r + ,..=n.
The equation/ (a?) = has in this case p roots each Oj,
q roots each O2, &c., the whole number of roots being
p + q + r+ ... =n.
* Proofs of this fundamental proposition have been Riyen by Gauohy,
Clifford and others *lhe proofs are however, long and diffioult.
THEORY OF EQUATIONS. 553
437. Relations between the roots and the coeffi
cients of an equation.
We have seen that if Oi, ag, as,... be the roots of the
equation / (a?) = ; then
f{x) = (^ — ai) {x — ttg). . ,{x — ar^.
Hence [Art. 260]
= a;«/Sfi.a?^^ + /Sf2.a?^2...+(l)~/Sfn,
where Sr is the sum of all the products of Oi, ag, a,,...
taken r together.
Equating the coefficients of the different powers of x
on the two sides of the above identity, we have
fi^i =  Pi, /Sj =p^,...Sr = ( iypr,...Sn = ( l^Pn
438. By means of the relations obtained in Art. 437, which give the
values of certain symmetrical functions of the roots of an equation
in terms of its coefficients, the values of many other symmetrical
functions of the roots can be easily obtained without knowing the
roots themselves.
The foUowing are simple examples :
Ex. 1. If a, 6, c be the roots of the equation
find the value of (i) Za^ and (ii) Za%^
We have a + 6+c= j),
hc + ca + ab=q
and abc=r.
Hence
a^ + b* + c^=(a + b + c)^2{bc + ca + ab)=p^2q.
Also,
S62c2= (bc+ca + abf  2abc {a + b + c)=q^ 2pr.
Ex. 2. If a, 6, c,... be the roots of x^+PiX^^+p2S^^+...\p^=0,
find the values of Xa^ and Za\
We know that 2a= i>i, Ikib^p^ and Sa6c=  p^.
Now (Sa)2=(a + 6 + c+...)*=2a2 + 2Sa6 [Art. 65];
Sa* = (Sa)2  2Sa6 =p^^  2p^ .
Again S^a . 2a = So^ + Sa^fc,
and Sa26 = 2a6 . Sa  3Sa6c.
554 THEORY OF EQUATIONS.
[For in Zab . Xa there can only be terms of the types a^b and dbc ; of
these the term a% will occur once, but the term abc will occur three
times, for we can take either a or 6 or c from Za and multiply by 6c,
ca or ab respectively from 2a&. Thus 2a5 . Za = Za% + 32a&6.]
Hence
2a»=2a2.SaSa5.Sa + 3Sa6c=(pi«2pj)(pi)>a(pi)3p8.
439. Theorem. If there are any n quantities Oi,
Og, Og, &c., and m be any positive integer not greater than
n ; then will
The following relations hold good :
tar = Xihtoi'^'^  Soi'^^Oa,
To prove the first relation it is only necessary to notice
that the product Xa^ . %a^^ can only, give rise to terms
of the types d^ and a^^Oi^ also every term of either
type will occur, and no term can occur more than once.
Thus 2ai . Soi"*^ = Soi"* + Soi"*^ Oj.
The other relations, except the last, will be seen to be
true in a similar manner.
Also, the product SoiOg'CWi . Soi can only give
rise to terms of the types a^a^.,,amri and aia^,.,a^\
the first of these terms can only occur once, namely as
aia2a8...a„i_i X tti; the second term will, however, occur
m times, for we get the term by taking any one of the
m factors it contains from 2ai and multipljring this by
the proper term of 2aia2...a„^i.
Hence
* %a^a2, . »(imi • 2cti = Stti^Osj^. . Mfn\ + wi . SaiOs. . .o^.
THEORY OF EQUATIONS.
555
From the relations [A], we have at once
+ m . Ztdi^h* • '^w [B].
If now Oi, ttg, as, &c. be the n roots of the equation
we know that
2^1 = — pi, 2aia2 = )2> 2aia2a3 = — Pa, &c.
Hence, by substituting in [B] and transposing we have
Soi"* + Pi . Soi^' + PaSai'"^ + . . . + Pmri Soi
+ p^.m = [C].
The formula [C] gives the sum of the mth powers of
the roots of an equation of the wth degree [m i^ n] in
terms of the coeflScients and the sums of lower powers of
the roots.
The sum of the mth powers of the roots of an
equation can therefore be obtained from the formulae
2ai+Pi=0,
2ai2+Pi2ai + 2p2 = 0,
2ai» HpiSoi^ + paSoi + 3p3 = 0,
Soa* + piSoi^ + paSai^ + ps . Soi + 4ip^ = 0,
If we eliminate Soi^ and Soi from the first three
equations we have
Pi Pi 3p3 + Soi'
1 Pi 2pa
1 pi
= 0; /. Xa,^ +
Pi P2 Sps
1 Pi 2p2
1 px
= 0.
To find Soi*" we must eliminate Xa^~^, Sai'""^..., Sai
from the first m equations, and we have
\
55«
THEORY OF EQUATIONS.
9l
p»
Ps • • • Pm—1
m.pm + ^Oi'
1
Pi
Pi "* Pm—2
(m  l)pmi
1
Pi ••' Pm—i
{m  2)p^^
• • 1
• • • • •
1 ...pm4
= 0.
0... 1 jpi I
The coefficient of ^Ui^ is a determinant of which all
the elements on one side of its principal diagonal are
zeros, the elements along the principal diagonal being all
equal to 1 ; the determinant is therefore equal to 1. Hence
Soi"* is equal to an integral function of p^^p^, &c.
If m be greater than n the relation corresponding to
[C] can be very easily obtained. For, since Oi, aa,..^ are
roots o{f(x) = 0, we have n equations of the type
Oi** +PiCh'^^ + i>2ai**"* + . . . +Pn = 0.
Multiply by Oi"*"^, Oa***"**,... respectively; then.we shall
have n equations of the type
ai"*
+ Piai"^^ + p./h'"^ + ' " +l>nai*'*^ = 0.
Hence, by addition, we have
t(h^ + PiXoT' '¥p,tar^^+ ... +prX(ir^ = 0...[D].
By means of the relations [C] and [D], which were
first given by Newton, it is easily seen that the sum of the
mth powers of the roots of any equation can he expressed as
a rational and integral function of the coefficients, m being
any integer.
440. Any rational and integral S3m[imetrical function
of the roots of an equation can be expressed in terms of
the coefficients by means of the relations
2ai = — )i, Xa^a2=p2, Xa^a2ai = — pz, &c.
Consider the sjTnmetric functions of the third degree.
i
THEORY OF EQUATIONS* 567
It is easily seen that
Thus we have three equations to determine Soi^, Xa^a^
and SoiajOa, and these are the only symmetrical functions
of the third degree.
Similarly each of the products p^, pip^y PiPsy pi and p^
can be expressed in terms of symmetric functions of the
fourth degree, and there will be as many smh equations as
there are symmetric functions of the fourth degree.
The same will hold good with respect to symmetric
functions of any other degree.
y^B
Transformation of Equations. 
441. We now consider some cases in which an equa
tion is to be found such that its roots are connected with
the roots of a given equation in some specified manner.
I. To find an equation whose roots are those of a
given equation with contrary signs.
If the given equation be f(oo) = 0, the required equa
tion will be /(— y) = 0. For, if a be any root of the
given equation so that/(a) = 0, then — a will be a root of
/(y) = 0.
Thus if the given equation be
p^Pif^ \ PiX'^"^ +i?2«^"' + +Pn = 0,
the required equation will be
'po(yT+Pi(yT^'+P2{yy'\ +i)n = o,
or poy"" Pxy""^ +i)2y^'  + ( lYpn = 0.
II. To find an equation whose roots are those of a
given equation each multiplied by a given quantity.
Let f(x) = be the given equation, and let c be the
quantity by which each of its roots is to be multiplied.
558 THEORY OF EQUATIONS.
Let y^cXy or  = x) then/(J = is the equation
required. For, if a be any root of / {x) = 0, so that
/(a) = 0, oc will be a root of /[^) = 0.
Thus, if the given equation be
the required equation will be
P' {^j+p^ (!r+^^ (fr+ +^»=^'
or poy"" + picy""^ + p^y^"^ + ■^pnC' = 0.
The above transformation is useful for getting rid of fractional
coefficients.
Ex. Find the equation whose roots are the roots of
each multiplied by c.
The required equation is
We can now choose c so that aU the coefficients may be integers ;
the smallest possible value of c is easily seen to be 6.
III. To find an equation whose roots are those of a
given equation each diminished by the same given quantity.
Let f(x)=0 be the given equation, and let c be the
quantity by which each of its roots is to be diminished.
Let y^X'Cy or a? = y + c ; then f(y + c) = will be
the equation required. For, if a be any root oi f(x) = 0,
so that/ (a) = 0, a — c will be a root oif{y + c) = 0.
An expeditious method of finding f{y + c) will be
given later on. [Art. 47 L]
The chief use of above transformation is in finding
approximate solutions of numerical equations ; it can also
be used to obtain from any given equation another equa
tion in which a particular term is absent.
THEORY OF EQUATIONS. 559
Ex. Find the equation whose roots are those of a;' Sx^ 9a; + 5 =
each diminish^ by c, and find what c must be in order that in
the transformed equation (i) the sum of the roots, and (ii) the
sum of the products two together of the roots, may be zero.
The equation required ia f{y + c)= 0, that is
(2/+c)«3(y + c)29(y + c) + 6=0,
or 2/' + (3c3)2/2 + (3c26c9)y + c'3c29c + 6=0.
The sum of the roots will be zero if the coefficient of y^ be zero ;
that is, if c = l.
The sum of the products two together of the roots will be
zero if the coefficient of y be zero; that is, if c* 2c 3=0, or
(c3)(c + l)=0.
IV, To find an equation whose roots are the reciprocals
of the roots of a given equation.
Let f{a)) = be the given equation. Then the equa
tion /( ) = is satisfied by the reciprocal of any value of
X which satisfies the original equation.
This transformation enables us to find the sum of any
negative power of the roots of the equation f(oo) = 0, for
we have only to find the sum of the corresponding positive
power of the roots of the equation/ f  J = 0.
"'' 442. A reciprocal equation is one in which tha
reciprocal of any root is also a root.
To find the conditions that an equation may he a
reciprocal equation.
Let the equation be
p^"" + Piotf"^ + p^""^ + + Pn = 0.
Then the equation whose roots are the reciprocals of
the roots of the given equation is
p' sr+^' ©""+ P' Gr+ +^» =<>'
or, multiplying by a?",
560 THEORY OF EQUATIONS.
The equation last written must be the same as the
original equation, the ratio of corresponding coefficients
must therefore be the same throughout. Thus
Pn i>«i Pn2 Po'
From the first and last we have Pn=Po^y so that
Pn= ±i>o> whence it follows that the coefficients are the
same when read backwards as forwards, or else that all
the coefficients read in order backwards differ in sign only
from the coefficients read in order forwards. These two
forms of reciprocal equations are often said to be of the
first and of the second class respectively.
^ 443. The following important properties of reciprocal
equations can easily be proved.
I. A reciprocal equation of the first class and of odd degree has one
root equal to  1.
n. A reciprocal equation of the second class and of odd degree has
one root equal to + 1.
ni. A reciprocal equation of the second class and of even degree has
the two roots ± 1.
[These follow at once from Art. 87.]
IV. After rejecting the factor corresponding to the roots given in I,
n, ni, we are in aU cases left with a reciprocal equation of the
first class and of even degree.
Y. The problem of solving a reciprocal equation of the first class and
of even degree can, by means of the substitution x+x~^=yf be
reduced to that of solving an equation of half the dimensions. For
the equation may be written
ao(iB«*» + l) + ai(a;**>+a;) + ...=0.
Divide by x*»; then
ttQ (a;»» + x^) + ttj (a;»»i + x'^+^) + . . . = 0.
Now, if x + xr^=y, x^+x~^=y^2^,
and, from the general relation
a;« + aJ"= {x^^^ + x"*^) (x + x^)  (a;*^ + aj^+^j,
it follows by induction that x^+x"^ can be expressed as a rational
and integral expression of the nth degree in y.
!
THEORY OF EQUATIONS. 561
Ex. Solve the equation 6x^  25x^ + Six*  31aj2 + 25a;  6 = 0.
As in in, the expression on the left has the factor x^1 corre
sponding to the roots =t 1. Thus we have
6(a:«l)26a?(a:*l)+31a;2(a^»l)=0.
Hence the required roots are ± 1 and the roots of
6a:4  25x» + 37a;2  25a? + 6 = 0.
Divide by x^; then
6 (a^' + ^a)  26 ^aj + ^^ +37 = 0.
X ^ x^ ^
Hence 62/^25^ + 25=0;
5 5
y = ory=^.
Prom x+  = jr , we have a;=2 or ;r .
X 2 2
From a: +  =  , we have x=^ (5±\/ll).
OS 8 6 ^
Thus the required roots are ±1, 2, ^ , ^ (5±n/11).
A O
h
444. The method of dealing with other cases of trans
formation will be seen from the following examples.
Ex. 1. If a, &, c be the roots of the equation a^+px^ + qx+r=Oy find
the equation whose roots are be, ca^ ah.
Since 6c= — = , if we put y = , the three values of y cor
€L d X
responding to the values a, &, c of a; will be be, ca^ ah. Hence the
T
equation required will be obtained by substituting  for x in the
given equation, so that the required equation is
or . r^+pry + gy*+y8=0.
Ex. 2. Find the equation whose roots are the squares of the roots of
the equation a^ + px^ + ga; + r = 0.
We have x(3^+q)=ip (x^\r) ;
aj« (x2 + g)2 =2)« {x^ + r)*.
Now put y=x^y and we have the required equation, namely
y{y+q)^=P^{y+ry,
s. A. 36
562 THEORY OF EQUATIONS.
Ex. 3. If a, &, c be the roots of x»+pa5*+gx+r=0, find the equation
whose roots are a(b + c), 6 (c + a), c{a + b).
a(b + c)=a{pa); &c.
Hence, if we put y=x{px)iy will have the values required
provided x is restricted to the three values a^byc; that is provided
X satisfies the equation
x^+px^\qx+r=0.
Thus if we eliminate x between the given equation and the equa
tion
x*+px'\y=0,
we shall get the required equation in y.
Multiply the second equation by y and subtract; then {y q) x=r.
Now substitute for x in the second equation, and we obtain the
equation required, namely
r^+pr{y q) + y{y g)*=o.
EXAMPLES XLin.
1. If O], Og, a, be the roots of the equation a^+px'\q=0, find
the values of
(i) (aa + «s)(«3 + «i)(«i + «2) (ii) (a2+«82ai)(«s + ai2aj)(ai + a22a,)
(iii) Sai». (iv) Sai». (v) Soi*. (vi) I^^a^ (vii) I^^^a^.
(viii) S (a.^^  a^a^) (a^^  a^a^), (ix) {a^^  a^) (a^^  a^a{) (a,'  a^a^) .
W Si— . (xi) S . (xii) Sv^^ .
a. Find the sum (i) of the squares, (ii) of the cubes and (iii) of
the fotirth powers of the roots of the equation x*'\px + q=0.
^8. If a, 6, c be the roots of the equation x^+paP+qx + r=:0^ find
the values of
(i) (b + eSa) (c + a36)(a + 63c).
\b c a] \c a bj \a b cj
<"■> (J  li) (p  ^) {I  ^) •
p^. Find the sum of the squares and the sum of the cubes of the roots
of the equations
(i) a:314a; + 8=0. (ii) ar*  22ai2 + 84x  49 = 0.
THEORY OF EQUATIONS. 563
5. If a, &, c,... be the roots of the equation
find the values of
(i) 2a«.
(ii) Sa».
(iii) 2i.
(iv)2;.
(V) S^.
0. Find the equation each of whose roots exceeds by 2 a root of
the equation
««4a;2 + 8a;l = 0.
7. Find the equation whose roots are those of the equation
each multiplied by c, and find the least value of c in order that the
resulting equation may have integral coefficients with unity for the coef
ficient of the highest power.
8. If a, 5, c be the roots of the equation a;' + px' + 9a;+r=0, find
the equation whose roots are
/^(i) hc,.ca,ah, i^(ii) 6 + c, c + a, a + 6. ' ("^) 6^ » ^Ijl^ » a^^ •
(iv) a(6 + c), 6(cfa), c(a + 6), >^) h^ + c^,c^ + a\ a^ + h^,
(vi) hca\cah\ahc\
9. If a, &, c, d be the roots of the equation a:<+px5+ga?' + ra;+«=0,
find the equation whose roots are
(i) 6 + c + d,<kc. (ii) 6 + c + d2a, Ac.
(iii) 6« + c2+d«a»,&c.
10. Find the equation whose roots are the cubes of the roots of
the equation afi +px^ + ga; + r = 0.
/^445. In any equation with real coefficients imaginary
roots occur in pairs.
For, if a + 6V 1 be a root of/ (a?) = 0, a?  a  6 V^^l
will be a factor of f{x\ and therefore [Art. 193]
a? — a + 6 V — 1 will also be a factor, whence it follows
that a — 5 V — 1 is also a root of / {x) = 0.
Corresponding to the two roots a±h v—l of /(a?) = 0,
/ {x) will have the real qxwdratic factor [{x — of + 6*}.
36—2
564 THEORY OF EQUATIONS.
*^4}6. In any equation with rational coefficients quad
ratic surd roots occur in pairs.
For, if a + V& ^6 a root of f{x) = 0, hjh being irra
tional, x^a^ tjh will be a factor o( f{x), and therefore
[Art. 179] x^a^s/h will also be a factor of/ (a?), whence
it follows that a — hjh will also be a root of/ (a?) = 0.
Corresponding to the roots a ± sfh, f (x) will have the
rational quadratic factor {(x — a)* — 6}.
Ex. 1. Solve the equation ar*  2a:* 22x2 +62x 15=0, having given
that one root is 2+V3»
Since both 2+^3 and 2  ^3 are roots of the equation,
(x2V3)(^2 + V3),
that is x^ix + l, must be a factor of the lefthand member of the
equation. Thus we have
(a;«4x+l)(a?«+2x16)=0. '
Whence the roots required are 2=t^3 and the roots of
a?» + 2a;15=:0.
Ex. 2. Solve the equation 2x^  12x^ + 46x  42=0, having given that
one root is 3 + /^  5.
Since S^fJ 5 are roots of the equation
(x3V^)(a;3+V^)
must be a factor of the lefthand member of the equation, which
may be written
{(x3)»+5}(2x3)=0.
Whence the roots required are 3 db ,,J^5, ^ .
Ex. 3. Solve the equation afi4afillx* + 40a^ + llx^  4a;  1 = 0,
having given that one root is J2+^S,
If Ja+fjb be a root of any equation with rational coefficients,
^a and ^6 not being similar surds, then =t^a±^6 will all four be
roots.
Hence in the present case
(x  V2  V3) (a;  V2 + v/3) (a? + V2  n/3) (a; + J2 + ^3),
that is a;* 10x2+1, will be a factor of /(x). The equation may
therefore be written
(x*  10x2+1) (x2  4x  1) =0,
whence the roots are ±^2 ±^3, 2 ±^3.
t\
THEORY OF EQUATIONS. 565
Ex.4. Solve ar*a:«9a;»14a; + 8=0,
haying given that one root is  1 + ^S.
x + l^S is a factor of f(x) ; and therefore, as / (x) is rational,
the rational expression of lowest degree of which x + l/v/3 is a
factor, namely the expression (« + !)' 3, must be a factor of /(x).
Thus we have
{(a;+l)88}(a;4)=0.
Thus the roots are
4, 1 + ^3, l + «4^3, 1 + uPi/d,
where w is an imaginary cube root of unity.
o
447. Roots common to two equations. If the
two equations f(x) = and ^(a?) = have one or more
roots in common, f {x) and (j) (x) must have a common
factor, which will be found by the process of Art. 98.
Ex. Find the common roots of the equations
a:»3a;»10a; + 24=0 and a;86ar» 40a; + 192=0.
The H. c. F. of the lefthand members will he found to be a;  4.
Hence a; =4 gives the conmion root.
^ 448. When it is known that two roots of an equation
are connected by any given relation, these roots can be
found.
Ex. 1. Solve the cubic a:'3x'10j; + 24=0, having given that one
root is double another.
Let a and b be the two roots and let a =26.
Then, since a is a root of the given equation
a8_3a«10a + 24 = (i).
Also, since 5 is a root,
or a«6a240a + 192=0 (ii).
The factor common to the lefthand members of (i) and (ii) will
be found to be a 4. Thus a =4 and 6=2; the remaining root of
the cubic is then easily found to be  3.
Ex.2. Solve the cubic 2x^  ISx^ + 37a;  30 = 0, having given that the
roots are in a. p.
The sum of the roots is equal to three times the mean root,
566 THEORY OF EQUATIONS.
15 5
a suppose. Thus 3a= ^, whence a= ^r. Divide f(x) by the
factor 2x + 5, and the remaining roots are given by x^5x + 6.
Hence the roots are 2, ^ , 3.
In the general case suppose that a and b are roots
of the equation /(a?) =0 connected by the relation b=(l>{a).
Then /(«) = and f{4>{^)} = have a common root,
namely a ; and this common root can be found as in Art.
447. Thus a and (a) can both be found.
fix. (i). Find the condition that the roots of x^+px^+qx + r=0 may
be (i) in Arithmetic Progression, (ii) in Geometrical Progression.
Let a, 6, c be the roots in order of magnitude ;
(i) a+6+c=36; /. 6=.
Hence, as 6 is a root, we have
.(lyKi)'^' (!)=»•
whence 2]f  9pq + 27r = 0.
(ii) a6c = 6*; /. b=i/r.
Hence, as 6 is a root, we have
r+p(r)t + g(r)i + r=0,
^ whence p^=q^.
449. Commensurable roots. When the coefficients
of an equation are all rational the commensurable roots
can easily be found.
It is at once seen that an equation with integral
coefficients and with unity for the coefficient of the first
term cannot have a, fractional root.
For if T^ be a root of /(a;) = 0, ^ being a fi:action in
its lowest ternis, we have
©■^^(r^ ^""O
Multiply by 6""*; then all the terms will be integral
except the first which will be fractional [for a is prime to
■
I
THEORY OF EQUATIONS. 667
b and therefore a^ is also prime to 6], and this is im
possible.
Now, from Art. 441, ll., any equation can be trans
formed into another with integral coefficients and with
unity for the coefficient of its first term ; hence, from the
above, we have only to find integral roots.
Now it is clear that if a be an integral root oif{x) = 0,
so that iT — a is a factor of/ (a?), a must be a factor of the
term which is independent of x. Thus if we apply the
test of Art. 88 to all the factors of pn we shall discover all
the integral roots.
Ex. Find the commensurable roots of ar*  21 s^ + 42a; + 8=0.
Here the commensurable roots, if any, are factors of 8. Hence
we have only to test whether any of the numbers ±8, ±4, ±2, ±1
«re roots. It will be found that 4 and 2 are roots. Having found
two roots the equation can be completely solved ; for we have
(a;2)(a;4)(.'c» + 6a; + l) = 0.
Hence the roots of the equation are 2, 4,  3 ± 2i^2.
/..
EXAMPLES XLIV.
Solve the equation a:^\2a?l^x^22x + l=0, having given that
2+y^ is one root.
/a. Solve the equation Sx*  23a;* +72x 70=0, having given that
3 +x/^ is one root
r 8. One root of
/ 3a:°4a;*42a^» + {
is fjt»jbi find the remaining roots.
V/ft. One root of the equation
the equation
3a:«  4a;4  42a;3 + 56a52 + 27a;  86 =
equation 2a;«3a;Hoa;H6a;3_ 27a; + 81 = is
J2 + Ay 1. Find the remaining roots.
j^ 5. Find the biquadratic equation with rational coefficients one root
of which is a/3  ^5.
^ 6. Find the b iqua dratic equation with rational coefficients one root
of which is tJ2 + ,^3. *
'•*— 7. Shew that oi^2x^2x\l = Q and ar*7a;2 + l=0 have two roots
in common.
<■ 8. Solve the equation a;^4a;3 + iia;2_ i4^^10=0 of which two
roots are of the form a + ^<J 1 and a + 2^ J 1.
568 J^ THEORY OF BQP^TIONS. 
^ ' * ■: 9^ Find the condition that tfie roots of x^+pa^+qx+r=0 may be in
/ Harmonical Progression.
^ lO. Find the conditions that the roots of x* +px^ + qx^+rx+8=i0 may
be in A. p.
1/ 1 1. Find the roots of the equation rr'  3a;'  13a; + 16 = 0, having ^yen
that the roots are in a., p.
•/l a. Solve the equation a^+2x!^21x^ 22x + 40 = 0, having given that
the roots are in ▲. p.
^18. Find the commensurable roots of
(i) x^7aP+nxlB=0,
(ii) a;*x'13a;2+16a;48 = 0,
(iii) 8a;'  26a;2 + 34x  12 i= 0.
^14. Solve the equation 4x'32a;'a; + 8=0, having given that the
sum of two roots is zero.
1/1 5. Solve the equation a;*+4a;*5a;'8x + 6=0, having given that
the sum of two roots is zero.
16. Find the condition that the sum of two roots of the equation
a;* +px^ + qx' +rx+8=0 may be equal to zero.
17. Solve the equation x^  79a;  210 = 0, having given that two of the
roots are connected by the relation a=2j3+ 1.
18. Solve the equation 3a;' 32a;2 + 33a; + 108=0, having given that
one root is the square of another.
10. Shew that, if the roots of the equation
a;*+npa;*i + — fp^^ ga;»»2+ ... =0,
be in A. p., they will be obtained from j)+r J ^^ IJl by giving to r
* the values 1, 3, 5,... when n is even, and the values 2, 4, 6,... when
n is odd.
20. Find the condition that the four roots a, /9, 7, d of the equation
x^{p3?+qx^ + rx + 8=^0 may be connected (i) by the relation aj8=75, and
(ii) by the relation 0^+75=0.
21. Shew that, if four of the roots of the equation
oa;* + 6a:' + ca;2 + da; + « = 0,
be connected by the relation a + j8= 7 + 8, then will 4a6c  6'  8a'd= 0.
22. If a, 5, c,... be the roots of the equation
a;* + j)ia;*i +p^'^^ + . . . +p,» = 0,
prove that (1  a') (1  feS) (1  c*) . . . = .4' + B' + C  ^AB C,
where ^=Pn+Pn3+ ^ B=Pn^+Pn4+"f
and C=p^^+p^^^+...
/ THEORY OF EQUATIONS. 569
^450. Derived ftinctions. Let
then, if a? + A be put for x, we have
f{x + h) =po (x + A)~ + Pi (a; + A)~^ +p2(x + hy^ + . ., +;>«.
If now {x + hY, (x + hy'\ &c. be expanded by the Bi
nomial Theorem, and the result arranged according to
powers of h, we shall have
f(x + h) =f(x) + h {np^^^^ + (n  1) piX""^
+ (n  2)p^''^ + +Pni}
+ higher powers of h.
This expansion is usually written in the form
f(a: + h)=f(a>) + hf'(x) + ^f"(x) +
[The reader who is acquainted with the Differential
Calculus will see that the expansion o{ f(x + h) in powers
of A is an example of Taylor s Theorem.]
It will be seen at once that/' (x) is obtained by multi
plying every term of / (x) by the index of the power of x
it contains and then diminishing that index by unity.
It will also be easily seen that f^\x) can be obtained
from f (x) in a similar manner, and so for /"' (x\ <&c. in
succession. We shall however in what follows only be
concerned with /' (x).
Def. The function /' (x) is called the first derived
ftinction of/ (a?), the function /"(a;) is called the second
derived ftinctiion of/ (x), and so on.
Thus if / {x) =PqX* +PiX^ +p^^ +P3X +i>4,
/' (X) = ipaX^ + Sp^X^ + 2p^i +^g ,
/" (x) = 12p^^ + 6piX + 2p^ ,
570 THEORY OF EQUATIONS.
•
451. Theorem. If{x) he any rational and integral
function of xandf {x) he its first derived function, then will
where Oi, Oa, Og, are the n roots, real or imaginary, of
the equation fix) = 0.
We know that , C .
f(x)=po (x  tti) (x  Oa) (a? as)
Hence
/ (a? + A) = ^0 (^ — 0^ + ^) (^ — G^ + ^) (^ — «8 4 A)
The coefficient of h in the expression on the right is by
Art. 260 equal to po x (sum of all the products n — 1
together of the n quantities a? — Oi, x—a^, , a? — c^).
But f(xhh)=f(x) + hf (x) + higher powers of h.
Hence f^(x)=poX(8um of all the products n — 1
together of the n quantities x — ai,x — a^, ,a? — a^.
Hence f (ai) = ^^ •\'^^+
In the above the quantities ai,a^, , a^ need not
be all diflferent from one another ; but if aj occur r times,
and Oa occur s times, &c., we shall have
/'(a;) = ^/(l>+3^> +
a? — Oi x — a^
452. Equal Roots. We have seen in the preceding
Article that if ai, aj, , an be the n roots of the equa
tion/(a?)«=0, so that/(a7)=)o(^ — cti)(a? — as) {x — a^^i
then will /' (a?) = po x (sum of all the products n—1
together of the n quantities a? — ai, a? — aa, ,X''a^,
Now, if any root, for example ai, is not repeated, so
that the factor a? — ai occurs only once in / (a?), then the
factor a; — tti will be left out of one of the terms of /' (a?)
but will occur in all the others ; whence it follows that
f'{x) is not divisible by a? — ai. Thus a root of f(x)^0
which is not repeated is not a root of f (a?) = 0.
THEORY OF EQUATIONS. 571
If, however, r roots of the equation /" (a?) = are equal
to Oi, the factor x — Oi will occur r times in f(oo), and
therefore a? — Oi will occur at least r — 1 times in every
term of/' (a;), for every term of/' (a?) is formed from /(«)
by omitting one of its factors. Hence a root of f{x) =
which is repeated r times is also a root off^(x) = re
peated r—1 times.
We can therefore find whether the equation f(ai) =
has any equal roots, by finding the H. c. F. of / (x) and
/' (x) ; and if / (x) be divided by this H. c. F. the quotient
when equated to zero will be an equation whose roots are
the different roots of / (x) = 0, but with each root occur
ring only once.
Ex. 1. Find the equal roots of the equation
x^  5x^  9x2 ^ 81a;  108 = 0.
Here f{x)=x*5x^9x^\SlX'108,
/' {x)r=4j^ 15x2  IBx + 81.
The H.c.p. of /(x) and/' (x) wiU be found to be x26x + 9, that
is (x 3)2.
Since (x  3)^ is a factor of /' (x), (x  3)' will be a factor of /(x),
and it will be found that / (x) = (x  3)^ (x + 4).
Thus the roots of the given equation are 3, 3, 3,  4.
Ex. 2. Shew that in any cubic equation a multiple root must be
commensurable.
This follows from Art. 445 and 446, and from the fact that a
cubic equation can only have three roots.
Ex. 3. Solve the equation x«  ISx^  lOx^ + 60x  72 = by testing for
equal roots.
Here /(x)=x»15x510x2 + 60x72;
/' (x) = 5x^  45x2  20x + 60.
It wiU be found that the h.c.p. of/(x) and/' (x) is
x3x28x + 12.
If now we divide /(x) by x'x28x + 12 the quotient will be
x2+ X  6, and the roots of x2+x  6=0 are 2 and  3.
Thus the given equation has only two different roots, namely 2
and  3 ; and it will be found that /(x) = (x  2)^ (x + 3)2. Thus the
roots of/(x) = are 2, 2, 2, 3,3.
572 THEORY OF EQUATIONS.
453. Continuity of any rational and integral
Amotion of x.
Let pf^'^ + piX^"^ + p^'^ + +pn be any rational
and integral function of Xi arranged according to descend
ing powers of x.
Then each term will be finite provided x is finite ; and
therefore, as the number of the terms is finite, the sum of
them all will be finite for any finite value of x.
It can be easily proved that the first (or any other
term) can be made to exceed the sum of all the terms
which follow it by giving to a? a value sufficiently great ;
and also that the last (or any other term) can be made to
exceed the sum of all the terms which precede it by
giving to a? a value sufficiently small.
For let k be the greatest of the coefficients; then
PiX^^ + ... bp^ k(x^^ + ...1) kx^ k ^ ''
Now ^{x\) can be made as great as we please by sufficiently
increasing x.
We can prove in a similar manner that Pn/(Pni^+ ••• +Po^*) can
be made as small as we please by sufficiently oiminishing x.
Now suppose that x is changed into a? + A ; then we
shall have
/(^ + A)/(^) = ¥'(^) + S/"(^) + ,
where the coefficients f {x), f"{x\ &c. of the diflFerent
powers of h are finite quantities.
Then by the above, the first term on the right (or if this
term vanishes for any particular value of a?, then the first
term on the right which does not vanish for that value)
will exceed the sum of all the terms which follow it,
provided h be taken small enough. But the first term
will itself become indefinitely small when h is indefinitely
small. Therefore /(xih) — /" (x) can be made a^s small
as we please by taking h sufficiently small. This shews
that as X changes from any value a to another value 6
THEORY OF EQUATIONS. 573
f{x) will change gradually and without any interruption
from f{a) to /(6), so that f{x) must pass once at least
through every value intermediate to f{a) and f(Jb),
It must be noticed that it is not proved that f{x)
always increases or always diminishes from f (a) to / (6),
it may be sometimes increasing and sometimes diminish
ing as a? is changed from a to 6 ; what has been proved is
that there is no sudden change in the value oi f{x),
^ 454. Theorem. If f(a) and /(/8) have contrary
signs one root a/t least of the equation f (ic) = must lie be
tween a and fi.
For since /(a;) changes continuously from /(a) tof(0),
it must pass once at least through any value intermediate
to /(a) and/(/8); it therefore follows that for at least one
value of X intermediate to a and /8 it must pass through
the value zero, which is intermediate to /(a) and/(/8)
since f(o) and /(/8) are of contrary sign. Thus the
equation y* (a?) = is satisfied by at least one value of oo
which lies between a and /S.
For example, if f{x)=ix?4x + 2, then /(1) =  1 and /(2)=2.
Henoe one root of the equation a;^  4a; + 2=0 hes between 1 and 2.
455. Theorem. An equation of an odd degree has
at least one real root.
Let the equation be/ (5?)= 0, where
Then /(+oo) is positive, /(0)=)jw+i, and /(— oo)is
negative.
Thus there must in all cases be one real root, which
is positive or negative according as pm+i is negative or
positive.
456. Theorem. An equation of even degree^ the
coefficient of whose first term is unity and whose last term
is negative, has at least two real roots which are of con
trary signs.
574 THEORY OF EQUATIONS.
Let a:^ + piaf^^ + +p2n = be the equation, p^n
being negative.
Then/ (+00) is positive, /(0)=p2n, and/ (—00) is
positive.
Hence, as p^n is negative, there must be one real root
at least between + 00 and 0, and also one at least between
and — 00 .
C* 457. The following is a very important example.
To prove that if a^ 6, c, /, ^, ft he aU real the roots of the equation
(X'a){xh){xc)'f^{xa)^g^{x^b)h^{xc)2fgh=0,
will always be real.
We may suppose without loss of generality that a>b>c.
Write the equation in the form
{xa){(xb){xc)f'}{g^(xb) + hHxc) + 2fgh]=0.
By substituting + 00 , 6, c,  00 respectively for x in
{xb)(xc)f^
we see that the roots of the equation {xb{xc)f^=0 are always
real ; and if a and p be these roots, where a > j3, then a>b>c>fi.
Now substitute +00 , a, /3,  00 for ^ in the lefthand numb^ of
the cubic equation, and we shall have respectively the following
results
+ 00, {g^Jah + hJacY, + {g^Jb p + hjc p}\ oo.
Hence there is one root of the cubic between + oo and a, one root
between a and j3, and one root between j3 and  oo .
If, however, a=p the above proof fails ; but if a=/9, then
{x b) {x c) /2, must be a perfect square, whence it follows that
b=c and/=0.
The cubic equation in this case becomes
(xa)(xb)^{g^{.h^)(xb)=:0,
the roots of which are at once seen to be all real.
If a be a root of the cubic equation itself, there will be another
real root less than j3. Hence all the roots of the cubic must be resJ,
for the equation cannot have one imaginary root.
The cubic equation considered above is of great importance in
Solid Geometry, and is called the PiacrlTnlnaWng Cnbie.
■>
458. Theorem. If /(a) and /(/8) are of contrary
signSy then an odd number of roots of f{x) = lie between
a and /3 ; also iff (a) and f (/8) are of the same sign, then
THEORY OF EQUATIONS. 575
no roots or an even number of roots of f (x) = lie between
a and /8.
Let a, 6, c, , k be all the roots of the equation
y (a;) = which lie between a and /8 ; then
f(x) = (a? — a) (a? — b) {x — c) {x — i) (a?),
where ^ {x) is the product of quadratic factors (correspond
ing to pairs of imaginary roots) which can never change
sign, and of real factors which do not change sign while x
lies between a and /8.
Then
/(a) = (a — a) (a — 6) (a  c) (a—k)<f> (a),
and /(/3) = (/8a)(/86)(/3c) O8&)0(/3).
Now, supposing a > /8 all the factors a — a, a — 6, ,
a — k are positive; and all the factors ^ — a, ^ — b, ,
13 — k are negative ; also (a) and (f) (13) have the same sign.
Therefore if/ (a) and/(y8) have contrary signs there must
be an odd number of the roots a, 6, c, ,k. Also, if
/(a) and /(/S) have the same sign there must be no such
roots or an even number of them.
459. Rollers Theorem. A real root of the equation
f'(x) = lies between every adjacent two of the real roots
of the equation f{x) = 0.
Let the real roots of f(x) = 0, arranged in descending
order of magnitude, be a, b, c, ..., k. Then
f(x) = (a; — a) (a? — 6). . .{x — k)<f> (x),
where </> (x) is the product of real quadratic factors corre
sponding to pairs of imaginary roots and these quadratic
expressions keep their signs unchanged for all real values
o{ X,
Then
/(a? I \) = (a? — a + \) (a? — 6 + \) . . . (a? — A? 4 X)
X {<f> (x) + \<f> (x) + higher powers of \}.
=f(x) + X ^^^'^^ 4 ^^ + ... + '^ + // X r + &c.
•^ [x — a x x — K <p\x) )
[See Art. 45 L]
576 THEORY OF EQUATIONS.
•^ x — a x—b x — h 9W
Now all the terms on the right except the first contain
the factor a; — a, and that term is
{x — 6) (a; — c),,.{x — A) ^ {x).
Hence
f (a) = (a — 6) (a — c). . .(a — k)<f> (a).
So f'{b) = (b'a) (b  c)...(6  A) <^(6),
/'(c) = (c a) (c  6)...(c  A;) 0(c),
Now 0(a), 0(6), 0(c), &c. have all the same sign.
Hence as a > 6 > c. . ., the signs of /' (a), /' (6), /' (c), &c.
are alternately positive and negative. Hence there is at
least one root of /' (x) = between a and 6, one root
between b and c, &c.
460. Descartes' Rule of Signs. In any equation
f(x) ^Q the number of real positive roots cannot exceed the
number of changes in the signs of the coefficients of the
terms in f{x\ and the number of real negative roots cannot
exceed the number of changes in the signs of th^ coeffi/dents
of /( oc).
We shall first shew that if any pol3momial be multi
plied by a factor x — a, where a is positive, there will be
at least one more change in the product than in the original
polynomial.
Suppose that the signs of any pol3momial succeed each
other in the order +H h + ^ h— , in which there
are five changes of sign.
Then writing only the signs which occur we shall have
+ + f + + 
+ 
+ +  + + + 
Now we cannot wiite down the second partial product
for we do not know that all the possible terms in the
THEORY OF EQUATIONS. 577
polynomial are present ; but whenever there is a change
of sign in the first partial product it is clear that if there
is any term in the second row of the sa»me degree in a?, so
that it would be put under this term which has the
changed sign, it must arise from the multiplication of
the next preceding term so that the two terms would have
the same sign. Thus whenever there is a change of sign
in the first partial product that sign will be retained in
the addition of the two lines of partial products. The
number of changes of sign, exclusive of the additional one
which must be added at the end, cannot therefore be
diminished.
Hence the product of any polynomial by the factor
x — a will contain at least one more change of sign than
there are in the original polynomial.
If then we suppose the product of all the factors
corresponding to negative and imaginary roots to be first
formed, one more change of sign at least is introduced by
multiplying by the factor corresponding to each positive
root. Therefore the equation f{x) = cannot have more
positive roots than there are changes of sign in the
coefficients of the terms in f(x).
The second part of the theorem follows at once from
the first, for the positive roots of /(— a?) = are the nega
tive roots of f(x) = 0.
The above proof may be made clearer by taking as a definite
example the multiplication of x^ + 2x*  ar* + 4a;8 + 3a:  1 by x1.
The signs of the two lines of partial prodncts will be
+ + '  + + 
  +   +
+  +  +
^ In the third line the only signs written down are those under
the changes in the first line, which changes are all retained in the
final product. Hence no matter what has occurred in the intervals
the number of changes (exclusive of the one at the end) cannot be
diminished.
461. Descartes' Rule of Signs only gives a superior
limit to the number of real roots of an equation, but does
s. A. 37
578 THEORY OF EQUATIONS.
Dot determine the actual number of real roots. The
number of the real roots of any equation with numerical
coefficients can be found by means of Sturm's Theorem.
Before considering Sturm's Theorem we shall shew how
to find algebraical solutions of cubic and biquadratic
(quartic) equations in their most general forms. Abel
has proved that an algebraical solution, that is a solution
by radicals, of a general equation of higher degree than
the fourth cannot be found, although particular forms
of such equations can be solved, for example any reciprocal
equation of the fifth degree can always be solved.
EXAMPLES XLV.
1. Solve the foUowing equation s each of which has equal roots :
(i) x812a;216a;4=0,
(ii) fl5*6a;« + 13x«24x + 36=0,
(iii) 16x*24x« + 16x3=0,
(iv) 2x*23a:S + 84x2~80a;64=0.
2. Find the condition that the equation aa:^+Sbx^+Scx + d=0 may
have two equal roots.
8. Shew that, if the equation aa^ + Sbx^ + Sex + d=0 have two equal
roots, they are each equal to
1 bcad
2 ac62
4. Shew that the roots of the equation
xa' xh' xc' xk'
are aU real.
5. Shew that all the roots of the equation
a2 62 c«
H a "• h ...=m + n^x ■
X a X fi X y
are real.
THEORY OF EQUATIONS. 579
6. If o^, Ag, Ag, .;., a^n be in descending order of magnitude, and if
& be positive, prove that the roots of the equation
{xai){xa^)...{xa^_^) + b{xa^)(xa^).„{x€uJ =
will all be real, and find their positions.
7. Prove that if a, 6, c, d be unequal positive quantities, the roots of
the equation
X XX , yv
+ — T+  +a; + d=0
xa xb xc
will all be positive ; and that, if roots be a, /3, 7, 9, then will
a2 62
(aa)(a/3)(a7)(a5) (6a) (6/3) (67) "(66)
= 0.
C2
(ca)(c/3)(c7)(c5)
8. Form the equation whose roots are the values of pu> + qur^^ where
b) is a fifth root of unity, and shew that the equation is
or*  6pq3^ + 6p2g2j. _p» _ g8 = 0.
O. If a, j8, 7, 9 be the roots of the equation
«* + ipo:* + 6ga;2 + 4rx + a = 0,
form the equations whose roots are
(i) ap + yd, ay+pS, ad + Py,
(ii) (a+/3)(7 + 5), (a + 7)(/3 + 5), (a + 5)(/3 + 7).
10. If a, j8, 7, 9 be the roots of the equation
x* + 4px^ + Qqx^ + irx + S=0
form the equation whose roots are
(a + /375)a, (a/3 + 75)2, (a/37 + 5)a.
11. If a^, Ajt a, be the roots of
x^ +PiX^ +P^+Ps=^*
shew that ^o^^^^^—^Pi^sPiP^+PiPs'
37—2
580 THEORY OF EQUATIONS.
Cubic Equations.
462. The most general form of a cubic equation is
a? + aa?^ + 6iz; + c = 0.
We have however seen [Art. 441, ill.] that by in
creasing each root by ^ , the equation will take the simpler
form a? hpx + q = 0.
We shall therefore suppose that the equation has
already been reduced to this simplified form.
463. To solve the cubic eqmition a^ •^px\q=X).
The solution is at once obtained by comparing the
equation with
a^  Sdbx + a^ + 6« = 0,
i.e. (x^ a{b){x + (oa + co^b) (x + (o^a f cob) = 0,
where o) is an imaginary cube root of unity [Art. 139].
Thus the roots required are
— a — 6, — (oa — 6)^6, — or^a — cob,
where a and b have to be determined from the equations
p = — Sab, q = a^ + b\
Whence a' and b^ are given by
2^V V4"^27.
464. The foregoing solution is a slight modification of
that called Garden's solution. It is a complete algebraical
solution of the equation and the values found for x would
satisfy the given equation identically. If, however, nu
merical values be given to p and q, the numerical values
of a and b cannot be found when ^ + ^ is negative, for we
THEORY OF EQUATIONS. 5^1
cannot reduce an expressiola of the form (3 + SV — 1)^, for
example, to the form a + ^s/ — 1. Thus when p and q are
numerical quantities such that j" + 9I7 is negative, Garden's
solution altogether fails to give a numerical result.
This case is called the 'irreducible ca4ie/ and
we shall see further on [Art. 466, Ex. 3] that when
^ + ^ is negative all the roots of the cubic are real.
It should also be noted that in any case the approxi
mate values of the real roots of a cubic can be obtained
much more easily by Horner's general process [Art. 473]
than by Garden's solution.
Ex. Solve the cnbio equation o:^ + 4a; + 5 = 0.
Comparing with a^  ^dbx + a' + &* = 0,
we have 8a6=4 and a* + 6'= 6,
whence a and h are given by
{54^1^}'.
The approximate values of a and 6 can therefore be found, and then
the roots are
— a — 6, (aa — oj^bt — <o^a — tab.
In this example the solution can be obtained in a very simple
manner. For, using the test given in Art. 449 for commensurable
roots, we are led to find that 1 is a commensurable root, and writing
the equation in the form {x \l){x^x + 5) =0, the roots are at once
seen to be
1, (i±VrT9).
Biquadratic Equations.
r
465. Several methods of solution of a biquadratic
equation have been given. In all of them the solution is
shewn to follow from the solution of a (Mhic equation.
The simplest method of solution is that due to Ferraii.
582 THEORY OF EQUATIONS.
To solve the equation oci^ \ pa:^ + qa^ +rx {■ s = 0,
Ferrari's Solution. Add (ouc+iS)^ to both sides of
the equation ; then
a^^paf^ + (q^a^)af' + (r + 2a^)x + s + ^ = (cuc + /3y.
Now the lefthaAd member will be a perfect square,
namely («^ + 1 ^ + N > provided
2\ + ^ = g + a2, jp\ = r+2a)8 and \^ = 8 + /3^.
Eliminating a and fi, we have a cubic equation to
determine X, namely
4(X«5)^2X + ^'3)(i)Xr)^ = 0.
One root of this cubic equation is always real, and if
this root be found the values of a and )8 are determined.
We then have
whence ar* + a? + X ± (ow; + )8) = 0,
where a, fi and X are known. Thus the biquadratic
equation can be completely solved.
Ex. Solve the equation
Add (cur+ j8)2 to both sides ; then
ic*+6«» + (14 + a3)a;a+(22 + 2ai8)x + 6 + /3«=((u; + /3)3.
The lefthand member is the square of x^ + Sx + \ provided
9 + 2X=14 + a2, 6X=22 + 2a/S and \^=5+p^.
Whence (X^  5) (2X  6)  (3X  ll)^ = ;
X87X2 + 28X48 = 0.
The real root of the cubic is 3.
THEORY OF EQUATIONS. 683
Then, taking X=3, we have
Hence {x^ + Sx + S)^= (x ~ 2)a,
whence we obtain the roots
466. Sturm's Theorem. Let/(a?) = be an equa
tion cleared of equal roots ^ and let/i {x) be the first derived
function oi f{x). Let the process of finding the highest
common factor oi f{x) and/i(a?) be performed with the
modification that the sign of every remainder is changed
before using it as a divisor, and let the operation be con
tinued until a remainder is arrived at which does not
contain x (this will always happen since f{x) = has
no equal roots and therefore f{x) and f (x) have no
common measure in x), and change also the sign of this
last remainder.
Let f{x), fz(x\..., fm{^) be the series of modified
remainders so obtained, of which the last, f^, (x), does not
contain x.
Then the number of real roots of the equation f{x) =
between a and )8, [^ > a] is equal to the excess of the num
ber of changes of sign in the series f{x), f(x)y fix),. . ,,fm(ai)
when x = a over the number of changes of sign when x = $,
For, let qi, ga**., ?mi be the successive quotients;
then we have the series of identities
/(^) = ?i/i(^)/a(^X
f (^) = qsf (os) f (x),
• • • *"* • • •
Now (i) it is clear that no two consecutive functions
can vanish for the same value of x, for in that case all the
succeeding functions, including fmQxi), would vanish for
that value of (x) ; and, (ii) it is also clear that when any
584 THEORY OF EQUATIONS.
one of the functions except /(a?) vanishes, the two adjacent
fuilctions will have contrary signs.
It follows from (i) and (ii) that so long as the increasing
value of X does not make f{x) itself vanish, that is unless
we pass through a real root of the equation f{x) = 0, there
can be no alteration in the number of changes of sign in the
series of Sturnis functions ; for no function will change
sign unless it passes through a zero value, and when this
is the case for any function, since the two adjacent func
tions have opposite signs, there must be one and only one
change in the group of three.
Next suppose that a is a real root of the equation
fix) = 0. Then /(a + X) =/(a) + \f (a) + &c. ; and as
f{a) = 0, the sign of the series on the right will, if X be
very small, be the same as the sign of + X/*' (a). Hence,
however small X may be, the sign of f{a — X) must be
opposite to that of /'(a), and the sign of /(a + X) must
be the same as the sign of /' (a).
Thus 05 X increases through a real root of the equatian
f{x) = Oy the series of Sturm's functions will lose one change
of sign.
Since we have proved that as x increases the series of
Sturm's functions never lose or gain a change of sign
except when x passes through a real root of the equation
f{x) = 0, in which case one change of sign is always lost, it
follows that the excess of the number of changes of sign
when x = a over the number of changes when x = ^ must
be equal to the number of real roots of the equation which
lie between a and y8.
To find the total number of real roots of an equation
we must substitute — oo and + oo in Sturm's functions ;
then the excess of the number of changes of sign in the
series in the former case over that in the latter will give
the whole number of real roots.
Ex. 1. Find the number of the real roots of the equation
Here / (x) = t* + 4a;«  4a;  13,
/i(x) = 4(a;3 + 3x2l).
THEORY OF EQUATIONS. 585
N.B. We may clearly multiply or divide by positive numerical
quantities as in the ordinary process for finding h.c.f. It will be
found that
f,^(x) = x^ + x + 4,
f^(x) = 2x + B,
f,(x)=19.
Substitute  oo , 0, +00 in the above functions, and the series of
signs will be
+  + ;  + + ; + + + + .
Thus there is one real root between  00 and 0, and one real root
between and + 00 .
Ex. 2. Find the number and the position of the real roots of the
equation a;*' 5a; + 1 = 0.
Here f{x)=x^5x + l^
and /i (a;) = 5 («*!).
It wiU be found that
/2(a;) = 4a;l,
/3(a;)=+255.
The following are the series of signs corresponding to the values
of a; written in the same line
00,

+
 +
2,
—
+
 +
1,
+
 +
0,
+
—
 +
1,
—
+ +
2,
+
+
+ +
Hence there is one real negative root between  2 and  1, one
positive root between and 1 and another between 1 and 2, the
remaining two roots being imaginary.
Ex. 3. Find the condition that all the roots of the equation
a:^+px + q=0
may be real.
f(x)=x^+px + q,
fi{x) = Sx^+p.
The other functions will be found to be
f^{x)= 2px~'6qj
Mx)=(27q^ + ^p^).
586 THEORY OF EQUATIONS*
The signs for  oo and + oo are
, +, +2i>, (27g* + V),
and +, +, 2p, ~(27q^ + ip»).
In order that the roots may be all real, it is necessary and
sufficient that there shall be three changes of sign in the first line
and none in the second, the conditions for which are that p and
27g* + 4p^ must both be negative, the second of which implies the
first.
(\
467. Although Sturm's Theorem completely solves
the problem of determining the number and the position
of the real roots of an equation, it is often a very laborious
process. In some cases the position of the real roots can
be determined without difficulty by actual substitution ;
and sometimes the necessity for using Sturm's Theorem
can be obviated by some special device.
Ex. 1. Find the number and position of the real roots of the equation
a;4_ 4x3.24. 40a. + 126 = 0.
Substitute in f{x) the values 1, 2, 3, 4, 5, 6 in succession, and
the signs will be+, +, , , , +. Hence there is one root (at
least) between 2 and 3, and one (at least) between 5 and 6 ■; but by
Descartes' Bule of Signs there cannot be more than two positive
roots.
Hence there are two positive roots which lie between 2 and 3 and
between 5 and 6 respectively.
We can find in a similar manner that there are two negative roots
which lie between  1 and  2 and between  6 and  7 respectively.
Ex. 2. Find the number and position of the real roots of the equation
In this case we should easily find the two negative roots which
lie between and  1 and between  4 and  6 respectively. The
positive roots would, however, probably escape notice (unless Sturm's
Theorem were used) as they both lie between 2 and 3 ; it wiU in fact
be found that/ (2) is +,/(2J) is , and/ (3) is +.
Ex. 3. Find in any manner the number and position of the real roots
of the equation
x^5jfi7x^ + 8x + 20=0.
By Descartes Rule of Signs we see by inspection that there
cannot be more than two positive roots and there cannot be more
than two negative roots.
Now /(I) is +, /(2] is  ; thus one real root lies between 1 and
2. Since / (oo ) is + , tnere must be another positive root which is
easily found to lie between 5 and 6.
THEORY OF EQUATIONS. 687
Change x into  x, then the negative roots of the given equation
are positive roots of
x^+5ixfi7x^Sx + 20=0.
Now f{x) must clearly be positive for all positive values between
Oand 1; and if x>l,
f(x)> 6a!* 15x^ + 20,
which is always positive since
4x6x2016a>0.
Hence there can be no real negative roots.
^ 468. We shall conclude by shewing how to find the
approximate values of the real roots of any equation.
This can be done in various ways; we shall, however, only
give Homer's method. We must first give the explan
ations of the separate processes which are employed.
469. Ssrnthetic Division. Suppose that when
is divided by iz; — \ the quotient is
and that the remainder is R, where R does not contiain x.
Then f(x) = Qx{xX) + R.
But Q X (a?T \) + iJ is at once seen to be
Equating coefficients of the different powers of x mf(x)
and in the expression last written, we have
bo^ao, 6i — X6o = ai, b^ — ^bi^a^,
From the above relations it will be seen that the values
of bo, 6i, 62, &c. can be obtained at once by the process
indicated below :
a© Oi Og ai ani an
Xbp \bi Xbz X6yt— 2 Xhn—i .
588 THEORY OF EQUATIONS.
First 6o = «o; multiply 6© by X and add to «i, the sum is
61; multiply 61 by \ and adil to •a* the sum is ig; proceed
in this way to the end.
Ex. Find the qaotieut and the remainder when
is diyided by x  2.
l_6 + 2 + 15 + 0+ 7
2812 + 6 + 12
146+ 3 + 6 + 19
Thus the required quotient is
the remainder being 19.
The above process is called the method of Ssrnthetic
Division. The method can easily be extended to the
case when the divisor is a multinomial expression, but
this extension is not needed for our present purposes.
470. The actual values of &o> ^i> ^2* ^^* ^ terms of Oq, 0^ a,, <&o.
and X can be at once written down ; they are
^o==^o> &i=ai + Xao, 62=^ + ^% + ^*^»
Kl = «nl + ^«n2 + ^^^n8 + ^'«n4 + • • • » + ^*~'«0 '
and Ji=a^ + Xa^_i+...=/(X).
Thus i^'. = aox**! + (oj + Xao) x""^ + ...
From the above we can obtain the formula of Art. 439.
For, if a, 6, c, ... be the roots of the equation /(a;) =0; then
•' ^ ' xa xb
= {a^x^"^ + («! + aao) x^~^ + (02 + 001 + a^ao) x^~^ +...}
+ { aoa;*»i + (oi + ftoo) a;*»2 + (Og + feoi + 6X) ^**~* + • • 4
+ ...
= woqX**"' + (noj + OqSo) a:*~*
+ (WO2 + OiZo + 0(,2o2) a:"8 + . . .
But /' (x) = noox**! + (n  1) aia;"^ + (o  2) OyT^^ + . . .
THEORY OF EQUATIONS. 589
^Equating the coefficients of like powers in the two expansions,
we*have
naQ=naQi
{nl)ai=nai + aQ'Sa,
(to  2) aj = TOOg + ^i2a + ^o^^'*
Whence the required result follows at once.
471. We have already seen [Art. 441, iii.] that in
order to diminish each of the roots of the equation /(a?) =
by \, we have only to substitute y + Xtorx \nf{x).
Let the equation whose roots are those of
a^"^ + (hx^^ + a^^ + ... + an = 0,
each diminished by X, be
Then, since y = x — \, the last equation is equivalent to
bo(x\)'^ + bi(x\y^^¥ ... +6ni(^^) + &n = 0.
The equation last written must be identical with /(a;) = 0.
Hence we have identically
f(x) = bo(x X)** + 6i (a?  X)^^ + . . . + 6ni (a?  X) + 6n.
From the form of the righthand member of the above
identity, it follows that if we divide /(a?) by x — \y and
then divide the quotient by (a? — X), and so on, the suc
cessive remainders will be the quantities bn, 6ni»"> ^i> h*
Ex. 1. Find the equation whose roots are those of
x*2a;8+3a;6=0,
each diminished hy 2.
Using the method of Art. 469 to perform the successive divisions,
the whole operation is indicated below, the successive remainders
being printed in black type.
1_2+ 0+ 36
2 6
1
3 + 1
2
4
8
1
2
4
11
2
8
1
4
12
2
1 6
590 THEORY OF EQUATIONS.
The first division gives the quotient a^+S with remainder 1 ; the
second division gives the quotient a^+2x+4 with remainder 11 ; the
third division gives the quotient a: +4 with remainder 12, and the
last division gives the quotient 1, and remainder 6.
Ex. 2. Find the equation whose roots are those of
each increased by 3.
The divisor is here x + 3, and the work is as under.
1 11+4 111+4
 3 + 1233
1 4 + 1129
 3 + 21
1 4 + lli29
1 7 +32
110
1 7 + 32
 3
110
Thus the transformed equation is
a^10a;2 + 32a;29=0.
We shall in future write the operation as on the right, the multi
plication and addition being performed mentally, and the result only
being written down.
' 472. In order to multiply all the roots of the equation
a^ + aios^^ + a^""^ + ... + ^n =
by ten, we must substitute ^ for a; in its lefthand
member. If we then multiply throughout by 10**, the
transformed equation will be
ttoic" + lOoaX^^ + lO^a^""^ + . . . + lO'^a^ = 0.
Thus in an equation with numerical coefi&cients the roots wiU be
multiplied by 10 by afl&xing one nought to the coefi&cient of x" \ two
noughts to the coefficient of a;'*"*, and so on.
For example, the equation whose roots are those of
each multiplied by ten, is
X*  20x» + 5000a; + 80000 = 0.
473. Homer's method of approximating to the
real roots of equations with numerical coefficients.
Having found (by trial or by Sturm's Theorem) two
consecutive integers between which a real positive root of
THEORY OF EQUATIONS. 591
the given equation must lie, the first step is to diminish
all the roots of the given equation by the smaller of those
integers. Then, by supposition, the transformed equation
will have a root between and 1. We now multiply
all the roots of the last equation 6y 10 by the process of
Art. 472, so that this new equation has by supposition a
root between and 10; now find by trial between what two
integers less than 10 the root lies, and diminish the roots
of the equation by the smaller of these integers. Then
again multiply the roots by 10, and continue the process
until the required degree of accuracy is attained.
After the roots of the given equation have been di
minished by the integral part of the required root, the
roots are multiplied by 10 in order to avoid decimals in
the work, the next integral root found must therefore
originally have been so many tenths. After again multi
plymg the roots by ten, the next integral root found
corresponds to hundredths in the original equation; and
so on.
By the above process it is clear that we are con
tinually approximating to the root sought; care must,
however, be taken that we do not pass beyond the root,
which would be shewn by the change, in sign of the
constant term.
The negative roots can be found approximately in
similar manner after changing x into —x.
a
Ex. 1. Find to two places of decimals, the positive root of the equation
x83a;4=0.
There can only be one positive root, and by trial this mnst lie
between 2 and 3. First diminish the roots by 2, and the transformed
equation wiU be found to be a:3+6a;^+9a;2=0. Multiply the
roots by 10 and we have the equation a^ + 60x* + 900a; 2000=0,
which will be found to have a root between 1 and 2. Diminish the
roots of this last equation by 1, and the transformed equation will
be a;» + 68a;2 + 1023a: 1039=0. Multiply the roots of this equation
by 10, and the resulting equation will be found to have a root
between 9 and 10. Diminish the roots of the last equation by 9,
and the resulting equation is «»+667a;«+ 113883a: 66541 = 0, which
could be used to obtain a more accurate result.
592
THEORY OP EQUATIONS.
The work is written as under, lines being drawn to indicate the
completion of each stage of the process.
3 4 (219...
2
4^
60
61
62
J"
+1
^000
900
961
1039000
102300
108061
 66541
630
639
648
113883
667
Ex. 2. Find the cube root of 30.
We have to find the positive root of the equation a;*  30 = 0. We
proceed as under
1
3
6
9
_B0 ( 3107
 3000
90
91
92 _
9300
9307
9314
2700
2791
 209000000
28830000
28896149
6733967
28960347
9321
It will be seen that after two or three multiplications of the roots
by 10, the numbers in the two last columns will become very much
greater than in the others; a contracted process can then be em
ployed, namely, instead of affixing one, two, three, <&o. zeros to the
coefficients after the first in order from left to right, we may cut off
one, two, three, <&c. figures from the coefficients after the first in
order from right to left. Proceeding in this way with the above
example after the stage at which it was left, we have
1 93!^1 2896034^
^^ 2896!^!^
 6733957 ( 31072325
 941617
 72661
 16727
 2247
The first of the new figures is 2 ; and after finding 2, the nnmbers
standmg in the columns will be 93, 2896220,  941517, the original
first column having previously disappeared. We then cut off one figure
from the second column and both from the first ; we then have only
to divide 941617 by 289622, cutting off one figure from the divisor ^
THEORY OF EQUATIONS. 593
at each successive stage, as in the ordinary method of contracted
division.
474. Imaginary roots. The nnmerical values of the imaginaiy
roots of an equation can theoretically he obtained in the following
manner, but the work would, except in very simple cases, be very
laborious.
Ex. Find the numerical values of the imaginary roots of the equation
x8 + 3xl=0.
 Put a + 1/3 f or x in / (x), and equate the real and imaginary ex
pressions separately to zero ; then we shall have
a»+3al3aj8«=0 and 8a*/Sj8» + 3/3=0.
Bejecting the factor /9=0, which corresponds to a real root of the
given equation, we have by eliminating /3 the equation
8a» + 6al=0.
Now a must be a real root of the equation last written, and this
real root will be found to be  *16109
Then j8"=3(a2+l), whence /3 is found to be 175438.... Thus
the required roots are  16109... ±175438... >/^.
EXAMPLES XLVI.
1. Solve the following equations :
(i) «»12a?+66=0. (ii) x»9x\2S=0.
(iii) «»48ar 620=0. (iv) rc» 21a; 3440.
^(v) «»2«+6=0. (vi) «»6xll=0.
. 2. Solve the following equations:
(i) «*+2x»+14a;+16=0. (ii) a:*12x6=0.
(iii) «*12aj»+24x+140=0. (iv) 4a:*+4x»7ic»4a;12=0.
8. Apply Sturm's Theorem to find the number and position of the
real roots of the following equations:
(i) «»8a;+6=0. (ii) «»a:»83x + 61 = 0.
(iii) 2a^a;310a? + 8=0. (iv) «* 14x8+ 16a; + 9=0.
(v) a;*7ar»+3a:20=0.
4. Find all Sturm's functions for the equation s^ + SpaP + Sqx + r=i 0,
and hence shew that, if p^<^qt there must be two imaginary roots.
6. Prove that the roots of x^ +px^ + r = are all real if either jp or r
be negative, and •%>'r be greater than 27i^,
b. A. 38
594 THEORY OF EQUATIONS.
6. The coefficients of the algebraical equation /(a;)=0 are all
integers. Shew that, if /(O) and /(I) are both odd numbers, the equa
tion can have no integral roots.
y 7. Shew that one root of the equation a?  2a;  5 =0 is 2*09455148.
8. Find the real positive roots of the following equations, each to
4 places of decimals :
(i) a»7a;+7=0. (iv) «*+a!»4a:»16=0.
(u) «»8«40=0. ^fv) a;*14«>16a:49=0.
y (iii) a»6fl?+9a;8=0. (vi) «»2=0.
9. Find the number and position of the real roots of
(i) a:* + 2a;823xa24x+ 144 = 0,
and (ii) a:* 262:2 + 483. + 9^,0.
10. Prove that the equation
a;«7a:»+16a;8+8a;4=0
cannot have more than 4 real roots ; prove also that these roots must lie
between 8 and  5.
11. Solve the equation
2a;»  7a:* + 6a;5  llx2 + 4x + 6=0,
having given that the real roots are commensurable.
la. Find the equation whose roots are the square of the roots of
a:»8px*3(ljp)a?+l=0,
hence shew that the given equation has three real roots for all real
values of 7.
18. Prove that the equation
ar»3pa:a8(l2>)a; + l=0
has three real roots for all real values of p.
Prove also that, if these roots be a, /3, 7 then
i3(l7)=7(la) = a(l/3) = l,
or /3(la)=7(l/3) = a(l7) = l
14. Shew that the equation whose roots are the sum of pairs of
roots of the quintic sfi +jpx + g = is
a;io _ 3px6 _ iigx*  ^j^^x^ + 4pga;  g* = 0.
16. Prove that the equation
a* + 4aa:» + 6a V + 4aa; + 1 =
has no real roots unless
and that the equation has two real roots if 0? is between these limits.
THEORY OF EQUATIONS. 696
16. Froye that, if a be the root of the equation
«* + ax»  6a?«  005 + 1 = 0,
, . 1 + a
go also IS :; .
1tt
Prove also that the other two roots are
1 , l + o
— and = •
a lo
17. Prove that, if a, /3, 7, ... be the roots of
** +PiX*i hp^* +... +Pn, = 0,
then 2a«/3*7 =  p^p^ + BpiP4  Bp^ .
18. Shew that, if o^, a,,. ..be the roots of
then Iai*a2*a^^a4+5q^+6(i0p'^=0.
882
ANSWERS TO THE EXAMPLES.
EXAMPLES I.
1. 4x.
6 6 ^ "5
e. 2ffi>+2ffin+2}i^.
8. a«6+106».
13 . 8
la. 5a*+ 3a»6  Safe^ +564.
14. hc+ica + iab.
a. 2«6y42f.
4. a*. 5. a?y4y«.
7 3a*+26'+c' + a64ac + 6c.
9. 2a+664c.
11. a:»«9.
18. 2x'^lxy+7y\
16. 3a'+26»8c»+6c+ca+a5.
16. «y. 17. 6y8«. 18. 2«+2y.
18. h + d. ao. y. ai. 4a. aa. 3«+3y.
as. 4n+4m. a4. 20. a6. 20.
EXAMPLES n.
82
8. ic»l.
1. 2ai»6a«+2a2.
4. ««+y». *• «*!• •" y**'*
7. **a:?+i«4. 8. l+aV+a*ic*.
11. 2««10aH»+6«*22iB36««+5a; + l.
ANSWERS TO THE EXAMPLES. 597
la. 4aj«  10«5y + 10x*y^  21a:«y«  Bxhf* + 5xy^ + y^.
18. Ca« + lla«616a*6«+20a«6»29a«6*+16a6»86«.
14. 2a«aJ»  3a««»y2 + SaVy*  lla*a%« + 6a*a^ + 20axy^^  10y«.
16. 2a3a2+a»+6a*6a«18a«+44a742a».
16. a^+6»+c»8a6c. 17. a;»+y*+««3a:y«.
18. 8a» + 276»c»+18a6c. 19. a;«l.
ao. «»25%«. ai. «»2ajV+y®
84. a8+8a«63+48a«6«+128a<5«+266&8.
86. (i) a'+4&*+9cS+4a56ac12&c,
(ii) a*2a»6+3a*6»2a6»+6*,
(iii) 62^2 4. c9a2 + a'd^ + 2a«6c + 2a6«c + 2a6c9.
(iv) l4a:+10»«12«»+9«*.
(V) ««+2a:«+8aj*+4a:' + 8a^+2a;+l.
86. (i) a«+6» + c» + 3(a«6 + a6*+a*c + ac«+6'c + 6c*) + 6a6c.
(u) 8a3276«8c»36a«6+64a6«24a«c+24ac«64Wj
866c*+72a6c.
(iii) l+8aj+6a;«+7a!»+6a^+3«»+a^. 87. 8a?«.
80. «*2a«x2a«»a*. 41. 0. 60. Sa«3Sa6c.
EXAMPLES in.
1. «8y. a. a;«+4y«. 8 9x*  12a?y + 16y«.
4. 8»2y, «• 1 +«+«*+«» 4«*.
6. «*+x»y+aV'+^'^y*' ^ l+2aj+3a:»+4a;»+6«*.
8. iii*+2m«n+3iii»n"+4jiiii»+6ii*.
8. l+2«+8x*+4a;»+6«*+6»'.
lO. l+a;*+«*+a:«. 11. l2aj+8a:«.
la. 28aj+2aj«. 18. 2jE«3icy+y'.
598 ANSWERS TO THE EXAMPLES.
14. x^'xy2x+y^+y + l. 15. a^+y^
16. x^2xy^\ 17. a + 2bBc.
18. a+26 + 3c. 1© Sa*+iab + h*,
ao. x^ky^+z^1. ai. a^^db+ae + b^hc + c^.
22, a*+46>+c«2a6+ac + 26c. 38. a+26 + 3c.
84. 9a«+4fi'+9c«66c+9<Ja+6at.
86. cx^+dajc. 86. Sojf  (36  4c) y. 87. a«3a6+&«.
88. a:*ay+y*, (a:+y)'«(a:+y)+««.
89. a»+«y+y*, (ac+y)^+2«(a?+y)+4z*.
EXAMPLES IV.
1. (a26)(a+26)(a2+462). 8. (2a;3a6)(2« + 8a6)(4ai» + 9a^i«).
8. (4+3a26)(43a+26). 4. (2y + aga;) (2/a5+a;).
6. 5ax(2ax + Sy){2ax~dy). 6. 4a2«»(3a;«+y«) (3«ay»).
7. 8(a5)(a + 5)(a*+62). e. 16(a6)(a + 6)(a»+6*).
8. 24a; (a; 1) (« + !)'. lO. 16a? (2  3a:*).
11. 46»(2a62)(4a« + 2a62 + 6*). 18. (a*46c)(a*2a«6c + 46«c2).
18. (a4)(a+2). 14. (4 a;) (3+ a?).
16. (l21a;)(l + 3a;). 16. 4(al)«.
17. oft (a 6) (a 35). 18. a«6(a+6) (a+46).
19. (6+<ja)(6+c6a). 80. (3a + 36cd)*.
81. (a;~2)(aj+2)(a:6)(a; + 6). 88. (5a;y) (6a?+y) (2a;y) (2a: + y).
88. (a«4y«««)«=(aj+2y2)«(aj2y«)«.
84. a«(a+&)(«^)(3a+&)(3a&)
86. (a:&)(a; + 62a). 86. («a)(a; + 2y+a).
87. (a+6 + cd)(a+6c + d)(a6+c + d)(a+&+cfd).
88. (x+y+a + b)(x+yab){xy + ab)(x+y+a'b).
ANSWERS TO THE EXAMPLES. 599
EXAMPLES V.
I. {x + l){X'l){x + a), a. {a + b)(cd).
8. (ab){c+d){ed). 4. {axhhy)(cx + dy).
6. (ax + b){cx^+d), 6. 2(ad) (a + 6 + c + cf).
7. (a + &)(a6)(a3 + a6 + 62). 8. (a6)«(aa + a& + 62).
8. (a+l)(al)(6 + l)(6l). lO. (a; + «) (««) (y + «) (y2).
11. («««  1) (y22  1). la (a: + y)(a: + «)(a;2x2: + 22).
18. («y)(a; + y + «). 1* (a; + 3)(a:3)(x«+2).
16. (a;« + 5a: + l)(a?25x + l). 16. (a:«+4a:2/ + y«) (a;*4a:y+2/2).
17. (x» + x + l)(x^x + l)(x*a?'\l).
18. (a;+a+&)(a?aft)(a? + a6)(a:a + 6).
19. (aj22yV)'. ao. («2ft + a6)(a:2aa^).
ai. (a;+6)(aj2 + a).
aa. {l'X^){l+y+x(ly)}{l+yx{l'y)}.
ad. (a;+y8«)(a;y+«). a4. (2yx+a) {y2xa),
86. (a86 + c)(a + 53c). 86. (2a 116 + 1) (a + 26 3).
87. (laaj)(l + aa; + 6a:2). a8. (laa;)(laajca;^).
89. (6c)(ca)(a6). 80. (6 + c)(c + a) (a + 6).
81. (a6)(ac)(6 + c).
88. (x^xy+y^){x(a+l)+y{h + l)}.
88. (a^ + a6)(ay2+65^). 84. (2xz){xy)\
86. (x*  y«) {y^  ««) («*  xy).
86. (x + 4)(a? + 2)(xl)(x3). 87. (a:+4) (x + 2) (a;« + 5flJ + 8).
88. x(xh5){x^+5x+10), 89. (a; + 2)(a; + 6) (a;2 + &B+10).
40. (x + 8) (2a; + 16) (2*2 + 35a: + 120).
600 ANSWERS TO THE EXAMPLES.
EXAMPLES VI.
1. '3 {y z)(z x) (x  y).
9. 5{yz) {zx) (xy) (x^ + y^+z^yzzxxy),
8. (6 + c)(6c)(c + a)(ca)(a+6)(a6).
4. (bc) (ea)(ab) {a+b + c).
6. (6c)(ca)(a6)(a8 + 6» + c3+62c + 6c«4.c«a+ca2 + a«6 + a&2
 9abc),
6. (6c)(co)(a6).
7. (6c)(co)(a&)[5«c* + c*a«+a262 + a6c(a+& + c)].
8. (6c)(ca)(a&)(a2 + 62^^^5g^g^^fl5j
9. (bc){ea){ab){a*+b^k<^ + h^e + b€^+(^a + ca^+a% + ab^
+ dbe).
lO. 24a6c. 11. 80abc{a^ + i^ + c^). 18. 4a6c.
18. 2a6c. 14. 4a&c. 16. 4(&c) (ca)(a&).
16. d{y + z){z + x){x+y),
17. 6(y+«)(« + a;) (a; + y) (a;2+ya+22^y;g+;ga; + a^).
18. (6c)(ca)(a6). 19. 2 (5c) (ca) (a6) (a + 6 + c).
ao.  (6  c) (c  a) {a  b) (3a» + 36a + Sc* + 66c + 5^^ + 5ab),
81. (6 + c)(c+a)(a + 6). 88. (6c) (ca) (a6) (a+6 + c)«.
88. (x+y+z)(yi? + «c + iC2/) **• {b + c){e + a){a + b){a+b + c).
86. 12x^2 (« + y + ^). 86. 3(6c)(ca)(a6).
87. 16 (6 c)(co) (a 6) (da)(d 6) (dc).
88. 27a«62(a + 6)8. 84. {a^+b^)^{c^+d^y,
86. (6  c) (c  a) (a  6) (a  <i) (6  <i) (c  d),
37. (6c) (ca)(a6)(ad)(6cf)(c<i) (6ccf + cda + <fa6 + a6c).
ANSWERS TO THE EXAMPLES. 601
EXAMPLES VII.
1. ah. a. 2a; 1. 8. a^yK
4. 2a;~7/. 5. «2y + 3z. 6. 4a^3a6 + 6^
7. a+26. 8. 2a:*3a; + l. •■ xa.
lO. a;2+a._6. 11. a»a; + 3. la. x23a; + 7.
EXAMPLES VHL
1. 12«* + 2air»  4a*a;2 _ 27a8a. _ iSa*.
a. (4a6)(a6)(3a8+l»«). 8. (x>2a? + 7)(6a;«+««44x + 21).
4. (a:»+5a; + 7)(7a:*40a:8+75x"40« + 7).
6. a;(« + l)(a? + 2)(a:2)(a; + 3).
6. a;(«l)(a;+2)(aj + 6)(x"2a?+4).
7. 2a (2a 6) (2a 36) (2a +36). 8. 6a; (a: +1) (a: 3) (a: 4).
9. (3a? + 2)(8a;8+27)(8a;827). lO, 3 (a;  3y)2 (a;«  4?/}.
11. (a;2y)(x8y)(aj4y).
14. {ac'  a'c)^ = (6a'  6'a)a (6'c  6c').
«*a
1.
4.
7.
9.
6a»a;»
a;Vl
4a;3j/«+l*
2xl
2x+3y
3x2^2*
EXAMPLES IX.
^ 3a6'c«a;«;2*
8. « •
a6
®' a + 46
aj2+a;y+y2
•■ x^xy+y^'
^ a;*a;y+y*
8.
(^
x^
1)«
3a; + l*
lO.
9a;«3a;2
6a:8+3a;»l'
18. l<yl)(«i:){«»).
ia ■
ANSWERS TO THE EXAMPLES.
ia> yi + ix + xy.
48
(I +3) (3
+ »)(.
+bH«
tsj
■■ «(••
a1(^
4.T
0. a«. 0. as. 1. ae. i.
d. as. a as. 1. «a 2.
a + b+e. aa. a' + b^ + c*+be + ca+ab.
(a+6+c)'. S«. a + bie.
(a+6 + BKa6+o)(a
+*
.)■
(s^y
a^i^»
(»p)t*~9)
41. 2.
(« + a)(i + 6)(i + c)
aateia + b + c)
—.. *7.
3(«+y+«)
EXAMPLES 3.
ft. «2fr.
•■ ■■J^si
..,;
ft, taa.
8. ±L
•. *1.
8.
S. I.
.6V«.
to. 6,
6i.
ANSWERS TO THE EXAMPLES. 603
50 a^c + b^a + c^bBahc
**■ 29* "" a^ + b^ + c^bccaay
14. [bc + ea + ab:i=^{b^c^ + c^a^ + a?b^'abc(a+b + c)}]^(a + b+c),
16. 6,7. 16. ±>/6.
, ^ 5
17. ±>/a6, ±^ab, 18. 0, ^.
//7i \
19. ±
N/d*")
ao. {a + b + €±^{a^ + l^+c^bccaab)}.
ai. a, V^— . aa. 2(^+6+c).
+ C
as. — =. a4. r — ^ .
a+o ao — ca
a5. — 5 r. ae. 0, a+6, — ^.
a9. 0, ±^^i^. so. 8,6. 31. 2, 3.
Sa. 1. 88. 0, a262. 34. a, 6.
86. T . ^ • «•• 0, 4 (a + 6). 87.  a,  6.
88. i(a6). 89. ±/s/a6. 40. 0, ±2^/^.
41. ±gv/3a»36^ 4a. ± >>/;^rr26?
48. 0. **• =*=a±6.
45. ^l{a + b+c±2^(a^ + b^+<^bccaab)}.
o
ab{a + b) Sab{a + b) ^ b^c^+(Pa^+aV)*
604. ANSWERS TO THE EXAMPLES.
^uoc
= =fc JL ji (be + ca + a5) (  6c + ca + a6) (6c  ca + a6) (66* + ca  a6)}.
2a6c^ ^
4
49. Yalnes between 3 and  ^ •
62. Yalnes between 3 and ^ .
68. X lies between  2 and 8, and y between  9 and 1.
54. X between  2 and 10, and y between  1 and 5.
a
55. 2
59. (i) a»a:«+(683a6c)a?+c»=0.
(ii) a^cx^ + «6 (62  3ac) + ac» = 0.
(iii) a;*6a;+ac=0.
eo. (1) 2^. (u) ^2. 66. .
EXAMPLES XL
1. ±2. ±V^2".
a. a, aw, aw*, 2a, 2aw, 2aw*.
8. a, —aw, aw*, 2a, 2a w, 2a w'.
4. l,i(l±V^O[^). 5. 0, 1, 3, 8.
6. 1, 2, (1±V^^) ^ 1, 6, i(7±3V6).
•• 5'T' ^('^'^^2). •. 3.1,1^2719.
*0. (l±is/^, i(a±N/?^.
ANSWERS TO THE EXAMPLES. 605
11. 0, 5, ^(6±V16). la. a, 9a, 4ad=aV^^.
13. 7a, 8a, (I±73i67). 14. 4,  6, i (  15 ± ^129).
21
15. 3, j^. 16. ±J(a + b), 17. ±a=fc6.
18. 2,i, i(3±V^. ^••'^'i' (l=tis/^).
ao. 1, j[l+V5±V{ + 2>/610}], [l^/5±V(2V510)].
ai. ±1, wi; ii^i^yri.
aa. 2, 2. .
as. 1,2,3,4. 34. ±1, i(7±3V5).
86. a, 6, c. ae. 9, 3±^47.
27. 9, 6, i(3±^/^^216). 88. a, 6, ^(a + 6).
89. a, 6; (a + 6)±l(a6)^/^.
80. a. 6, ^(a + 6), i(a+6)±(a6) V^.
81. a, 6, g(a+6). 88, a, 6, ^ {« + ft±^(a5) V^^.
88. a, 6, 2{a+6±(a6)Ay^}.
84. a, 6. 86. a, &, and roots of (a  x) (a;  6) = 16 (a  6)'.
87. a26, 62a, 5{a + 6±(a6) V^^15}.
88. Roots of «(a«) = r^/6=t A/l + gj .
606
ANSWERS TO THE EXAMPLES.
a« 6* I —
89. = 1 —^Jah,
b * a
41. 0, a + h.
40.
be ca ab
TT » "IT* T •
a* + f^ 2a5
+ 6 ' a + 6'
42. {6±^(6a+4)}, {a±V(a3+4)}. 48. i{a±^/(a»46«)}.
(a + 6 + c), 5(a + 6 + c)±iv(Sa3226c).
46. a + 6 + c, (a+6 + c)±v(Sa«7S6c).
46. a» &) c
47.
"'^Vl a + b + c^d r V I c + <i«6 J
EXAMPLES Xn.
1. a; = l, y= 1.
8. «=8, y=6.
8. a;=a, y=b.
11. «=3, y = 3, ;8=1.
18
8
a. ^=y,y=3
4. «=g, y=3.
e. x=a5, y= a 6.
8. x=y=a,
lO. a;=a(a6), y=fe(a6).
12. ^=2»y=3. *"6
18. x=y=2=l.
14. a;=6 + ca, y=c4a6i «=a+6c.
15. a5=6 + c, y=c + a, «=a + 6.
16. x=:(2a + 6 + c), y=i(a + 26+c), «= ^ (a + 6+2c).
17. a;=y=«= — . , . .
18. a;=i(2a+6 + c), ys^Ca + Sft + c), z=^(a+6+2c).
ANSWERS TO THE EXAMPLES. 607
a
**• * (a5)(ac)' ^"(6c)(6a)' '""(ca)(c&)*
ao. x=af y=bt z=c,
21. «= a + 6 + c, y=a6 + c, « = a + 6c.
9a. x=a(&c), y=6(ca), z=c{ab),
aa. a;=l, y = 0, ;8 = 0.
24. d; = a&c, y=6c + ca + a6, «=a + 6 + c.
a{ab){ac) » y »
^ , , a (a + b + c)
27. a; = 6 + c, y=c + a, «=a + 6. 28. g= . ^ ^, , ^, Ac
(a6)(ac)
29. x=2(m+n), 2^=^(n+0» « = 2(^+^)
80. ar=y=«=I2+m'+n'mnnZiwi.
81. Zar=my=fi« = l.
(a + a){a + b)(a + c) . ^^ 1 /r . . ^ « x ...
82. x= ^ / o\/ \ » &0' 83. a;=5(6 + c + d2a), <feo.
(a^)(o7) 3' "
84. a;=a&c(2, y= (6c(2+c(2a+<Za5 + a5c),
g=bc + ea¥ab+ad+bd+cd,
w= (a+6 + c+rf).
EXAMPLES Xm.
1. 12, 11. 2. 1, 1, 1, 1.
8. ^3,:.1;=.4^,^J^.
4. ±2, ±3; ±y^3, =Fy^/3.
6. 12,7; 7, 12. 6. a, 6; 2aft, 25a.
_ 2a6 2a6 a /= — s 6 i
60' 0+6*
608 ANSWERS TO THE EXAMPLES.
9. ±g, ±6; ±0, db.
6»_a» /6«a»
/6»a» /
3a
11. ±7, =f5; ±5, =f7. la. 9, 4; 4, 9.
18. 64, 8; 8, 64. 14. 6±^/30, 6=f>/30.
15. i(i=tV^nT), ^{l^J^Hi); 2, 1; 1, 2.
7 14
16. 1, 1; 2=1=^7, 2=Fis/7. 17. 2, 4; ^, y.
18. 2,1; 1,2. 18. 2a, 26; a, 6.
ao. 5,2. 81. &, a.
88. 6,6; (l±>/5), (1t>/5).
/~±ab , , / dbad
84. ±i^/l + o*, ±Jl + a«. 86. 8, 4; 2,4.
a
86. 4, 2; 2, 4; 3±,y^7d, StvA^^IS.
1 « ^ a+b b—a
89. 1,; 1,2. 80. 0,0;^,^.
81. 6, a; ~, — . 88. 6, a; 6, ; ^, a; J^^, y^^.
EXAMPLES XIV.
be ca ah be ca a6
a? _ y _ « __ , 1
ANSWERS TO THE EXAMPLES. 609
8. «(6+c)=y(c+a)=iS!(a+6)=± a/ ]2(^+<^) (<'+«) («+^)[ •
A o n. 2a6c 2abe 2<ibc
* ' * bc + ca+ab* bcca+db* bc+cadb*
^ ^ ^ 26c 2ac 2ab
5. 0,0,0;^ , — r, —Z — .
6 + ca c+a6 a+6c
6. x=y=z=s^2,
X V z 1
7. , = — ^^ = ~ — =±
b+.ca c + ab a + bc ^{2a + 2b + 2c)'
a u c
X y z 1
*' a« (63 + c«) ~ &> (<J«+a«) " c« {a^ +b^)^ '^ 2abe*
X V z
a(a+6 + c) 6(a6 + c) c(a+6c)
^{(a + 6 + c)(a6 + c)(a + 6c)}*
_ — —A * _ y _ z
= :t
V {(6+ca) (<j+a6) (a+6c)} '
X _ y i___ 2(a^» + 6» + c«)
13 ._,^ /(6+l)(<! + l)
• ^=^*V a + 1 '^
15. x^a±^^—^ l,&o.
17. 1, 2, 8. 18. 8, 6, 7. 18. 0, 4, 5.
ao. 8, 8. 4. 21. 0, 0. 0; , , .
a A, 39
610 ANSWERS TO THE EXAMPLES.
22. a, 6, c; a, 6, c; a, 6, c; (1±n/7), j(1±n/7).
'• ^=;^G+5)'*'
26
' (6 + c)2(c + a)(a + 6) (c + a)a(a + 6)(6 + c>
« 1
27. a, 6, c; (26 + 2ca), i(2c + 2a6), g(2a+26c).
28
20
x=±^:;— (6cca + a6) (6c + caa6), &o.
4aoc
(ca  6^) (oft ~ c^
. a, 0, 0; 0, 6. 0; 0, 0. c; A^—Li^.
/>/./> 3 a a b S. b c c 3
80. 0, 0, 0; ^a, ^^ 3; g' "2^' 2» 2' 2' "2^'
^(a + 6 + c), 2(«2» + c)» 2(* + ^'"^)
82. r — = — — = i =>/( «^<^)»
6c ca ab ^^
2X ^ by ^ cz _ I abc
^c" ca" a 6~ V 6c+ca+a6*
ax
b
EXAMPLES XV.
I. 20, 80. 2. A £10, B £16, C £25. 8. 20 years ago.
4. 2«. 5. 6,16,30. e. 6 days.
7. 1800. 8. 58. •. 30 miles.
10. 120 lb&. 11. 4 days. 12. 36, 9, 12, 15.
ANSWERS TO THE EXAMPLES. 611
18. 48 miles. 14. 16 miles. 15. 54, 81, 108.
17. A £460, B £226, C £237. 10*., D £87. 10«.
18. :t6. 10. 88,83. 20. 18 miles.
21. At 1 o'clock, 16 miles from Cambridge. 22. A £10, B £6, £1000.
28. 26. 24. 9, 7; 8^2,^2. 25. 60 miles.
26. 676. 28. 3 miles an hour. 29. 8 hours.
80. 268. 81. 2 gals, from the first, and 12 gals, from the second.
82. 16 minutes past 10. 88. 9 o'clock, 30 miles from Oambridge.
84. 46 and 22^ miles an hour. 85. £3.
86. 460 miles. 87. 30 miles.
EXAMPLES XVI.
8. a+&+c+a&c=:0. 4. (6+ca)(c+a6)(a+6c)=8.
8. a>+2c»8a5«=0. 15. l»+m«+n?Zmn4=0.
16. aP+6)ii'+cn'+Zmn=4a&c. 17. a'+&'+c'a6c4=0.
18. a»+6'+c'6a6c=0, y»«>+«»a^+»»y8+af»yV=0.
19. 26»c»=6an»>c«.
20. a» + 6«+c'6c(6+c)ca(c+a)a6(a+6)=0.
EXAMPLES XVn.
I. a* 6*. 2. 1. 8. ^^. 4. 1.
6*
5. afy. 6. a:*+l + ar*. 7. a!+y+«3a;iyizi.
8. af»af"^. 9. a^+a^ar^+a^ar^+aTnTayf +af*. lo. a:+y.
la. 4«»+3a; + 23xi, 18. a;* + 2xi+l+2a;i + af~^.
89—2
612
ANSWERS TO THE EXAMPLES.
20. (i) a^a*b^ + a^b^ah* + aib^b^,
(ii) a^x^a^x^yi+ah:^ya^x^yi{aix^y^y^.
(iii) a^+J^x^+<^x^bcxcax^(ibx^.
(iy) {x^ +y^ +z^ y^ z^  z^x^  x^y^] {{x+y \ zf
+ Bx^y^z^ {x+y{z)]+ 9x* y^ z^,
EXAMPLES XVm.
I. 2^3.
2. 6^/16.
8. ?s/2. 4. 52.
5. 0. e. UJ. 7.
2^3 + 3^2 ^/80 ^ ^/30+2^33^2
12
12
•. (^/21+^/10^/14^/16). lO. i(V6+N/10~V21V35).
^^^ 44^4 + 2,^2 + 4^ ^^ 3^3_^1^ ^3^ ^21.
14.
4/124^4
21
16+^10
15. 72^18.
17. 1+V6+V7. 18. V31
16. 6V3.
19. 2^8. 20. 1.
22. V2+V6a>/3. 23
• V 3
5 ' v ^ 3
24. 1+V2+V3. ae. V3+V2+V6. 26. 2+^2^/6^6.
EXAMPLES XIX.
1. 2a^3y'.
4. 6aa86«2c>.
7. 7 + 8a;« + 6a:8,
2.
5.
8.
X*  dxY'
8. a2&8c.
6. 2a;«2ay«y*.
x^x + 2x^ + xK9. —2^.
y Sac
ANSWERS TO THE EXAMPLES. 613
iO. x^2xixi, 11. a:i2a;i+x^. 12. oT^xix^a^,
18. xS, 14. x^xy^y^ 15. l8«» + 2aJ*.
16. 2(6c+ca+a6). 17. x^xiykz)yz, 18. a^+ft*.
21. ^=20, B=68, 0=44; or ^ = 62, B= 68, 0=76.
28. af^gh, bg=hf, and ch=fg.
EXAMPLES XX.
5. a;=:8. 8. a:&:c=2:8:4.
17. (i) 1 , 1. (u) 0, 1. (iu) 00 , 0.
EXAMPLES XXL
2. 2,4,6,8; 2, 4, 6, 8. 4. 2,4,8.
14. 6, ±12, 24, Ac. 17. 3, 9, 15.
EXAMPLES XXn.
1. 81. 2. (rl)(rl) ,1000.... 4. lib., 21b., 10241b.
7. 46. 8. 6. 18. 502 or 861. 15. 288, 289 or 290.
16. 2775 or 2525. 17. 135. 18. a=8, 5s0, c=6.
19. 7. ao. 1089. 28. 142857, 285714. 24. 166, 199.
EXAMPLES XXm.
1. (20/{4}». 2. 185. 8. 3"; [12/(4)».
1
2'
5. 260. 7. 571 (n1). 8. 5n(nl)im(ml) + l.
•. j{n(nl)(n2)m(wl)(m2)}.
614
ANSWERS TO THE EXAMPLES.
11. n(nl)(n2)(n3).
14. ^n(n4)(n5).
12. 3m^
IPH
18
17.
m + n2
• ml nl '
\mn
20.
g.,„ .,,.. .,. _.. 1^ 1^, ^^i « \in(\n)
28. 2(mn+m+nl).
24. 2Za+ 2Za5  2 (n  1), where n is the nmnher of giyen diameters.
m
EXAMPLES XXIY.
1. «»+6aas*+10aV+10a»a^+6a*af+a».
2. 32a»  80a*x + 80aV 40a«x» + lOoa?*  a?'.
8. l6x?+16a:*20x«+15«86ajio+«'^
4. 16a* 96a^+216a« 216a' +81a8.
5. 16»«96aj«+216a^216a?«+81.
e. «w10a;8y»+40a;V'80«V*+80a:V"32y".
7. 406«V.
3_89
lO. 
18. (ir
xh^.
' Eli
11. 701!*.
8W4V«.
•. 924i^.
12
21
l!i
8'*x'^V
14.
• [ion.
a;">and
^
liilH
eU
.a;*'^**«*
r nr r nr
15. {Sxy^  30 (3a?)i^ + 420 (3a;) V • •  946a?2 (2y)W + 45« (2sf)"  (3y)i».
16. 924c<<. 17. 643&p^ US5afi,
(l)*[2n/n[n. 28. 7 or 14. 24. 7.
1. Conyergent.
4. Oonyergent.
7. Diyergent.
8. Diyergent.
EXAMPLES XXVI.
2. Oonyergent. 8. Oonyergent.
5. Oonyergent. 6. Diyergent.
8. Oonyergent if x > 1 ; Diyergent if a; :f 1.
lO. Oonyergent if a; > 1 ; Diyergent if « 4* ^*
ANSWERS TO THE EXAMPLES. 615
11. Convergent itx:^!; Divergent if a; > 1.
12. Convergent iixi^l; Divergent if x > 1. 13. Divergent.
14. Divergent. 10. Convergent if m < 1 ; Divergent Hm^l,
16. Convergent if m < 1 ; Divergent if m ^^ 1. 17. Divergent.
18. Divergent. 19. Divergent. 20. Divergent.
21. Divergent. 22. Convergent if x < 1, Divergent if x > 1.
Iix=lf then Convergent if ft > 1 and Divergent if ft :f 1.
23. Convergent if a; < 1 ; Divergent itx^l,
24. Convergent if a; :^ 1 ; Divergent if re > 1.
26. Convergent if a; < 1, Divergent if a; > 1. If a;=l, then Convergent
if m < i and Divergent ifm'jfii,
27. Convergent if a? < 1, Divergent if a; > 1. If a;=l, then Convergent
if ft < i and Divergent if ft ^ }.
EXAMPLES XXVn.
1. (i){r + l)xr. (ii)i(r+l)(r + 2)aj»'. (jij) (^+^ (^+^^(^ + ^^) ay.
^^^'^ ^' 3.6.9...3r • ^^^ ' 3.6.9...3r ^'
(TiW i). g'2'l^'7...(3r8) 3.8 . 13...(5r2)
^^^ ^ ^^ 3.6.9...3r ^* ^^' 1.2.3...r *'
2.3.8.18...(5r7) . q (g+i?) (g + 2p)...(gt^^^. 1?) 
<^^ 1.2.3...r ^ • ^"^^ Fir *^'
(.)(2.r> ^^y^) (g^J,
^5.3.1.1. 3. 5...(2r7) ^^3^
^""^ 4,8.12...4r **•
2. (i) The ninth term, (ii) the eighth term.
8. The 39th term. 4. The first and second terms.
616 ANSWERS TO THE EXAMPLES.
5. After the 12th term. 8. (i) ^'^'1"}\ ^^ a^'^^xK
^' 3. 6. 9.. .or
(ii) 2a*****. (iii) 4ra*'a?*'. (iv) ' ' '"} a^x^ when n is
^ ' ' ^'2.4 .6...»
1 . <5 . o.,,{n — 2) _  1 • Ti
even, ^ . ^ — ) =( a"a;* when » is odd.
2 .4. 6...(nl)
(v) (2ra + 2r + l)a*iic**. (vi) (  1)** 16 (r  1) aS'****.
EXAMPLES XXVm.
1. (i)2. (ii) ^, (iii) 44^4. Civ) V272.
(V) 1. (vi) 4^4. (vu) ^A. (viii) i,
81. 245/8. 88. 246,792. 88. 462. 84. 85.
85. Coefficient of oe^'' is 88*"2»»^2 a»\ of a;^*^* is  B^^^ 2»» a3^»
and of a;**^^ is 0.
88. ^(« + l)(n + 2)(«+8)(n+4)(» + 5).
88. 2*+*^ 13n + rl / r_ [3>il .
EXAMPLES XYTT,
1 ^Q 8 • 4 8
8
6(rc + 6) 5(x + l)  a:8"^^*
3 5 1
• 4(a:+3) 8(x + 6) 8(x + l)'
4 ^ ^  H 2 1 ^ 1
' ' X (x + l)^' "• (a;2)'~i^^'*"a; + l*
6 ^ 7 13 1 . 4a;8
• 12(a;+l)~3(a;2)"*'4(a;3)* ^' 5 (x + 2)'^5^^iy
ANSWERS TO THE EXAMPLES. 617
1 1 1 1 1
•• .l10a;'*"8(l + 3a;)"8(l + 3a:)«' * 2{x^+l)^ 2(xl)^'
8 21 21 7
^®* 2 (lSx)* ■*■ 8(l.8a:)a "*" 82 (l8x) "*" 82 {1+x) '
1 _1 1_ 2 4 a;+2
**• a:»+l'*"«2 x + B' {x'2)^'^ 5(x2)'^ 6(x^+iy
2 11 llg4
*•• 6(a;l)«'*"26(xl)"25(a:«+4)*
8 1__ 1 a;+2
• 6(x2) 2(a;l)a"2(ajl) I0(a:2+1)*
JL i. 1 17 1 8
**• ac«~16aj"« + l'*"l6(iC + 2)"**(a; + 2)a'*"4(a;+2)»*
1 1 11 1 J 3_
*•• 4(aJ+2)»'*"6(x+2)a''"l44(x+2)'*"9(a;l)'"8a;» 16aj*
17. (ir{2*8Hi}. 18. (y'''J(8n+7).
20. g(8*l){{l)*l}. 21. i{9 + 6*«2.8»*+«2«+<}.
. (n« + 7»+8)2»»; i (n»+9»«+14n)2»»*.
EXAMPLES XX^IL
1. 1*262. 2. 148169. 8. £1146*74.
5. £742. 19«. &d. 7. £785. 10«.
8. £1979. 58, 6d. 9. £1785 nearly. lO. £122*58.
EXAMPLES ^^ynr.
1. i{(3n+l)(8nd4)(8n+7)(8n+10)1.4.7.10}.
8 18.7' (4n+3)(4n+7)) ' *'"168'
8. :irn(n+l)(8n«+23n+46). 4. n(n+l)(»f2)8«.
618 ANSWERS TO THE EXAMPLES.
d. i«(n+l)M»+2). 6. ^n(n+l)(n+2)(3n+6).
7. i(2nl)(2i»+l)(2n+8)(6» + 7) + g.
jl 6nHl Q 11
■• 180 12(2n+l)(2»+3)(2n+6)' *«"*180'
^ 8n+5 __6
36 6(n+l)(n+2)(n+8)* *""86*
5 2n+5  __6
*^' i 2(n+l)(n+2)' ^«i'
• 8 8(2n + l)(2n+3)* «''"8*
29 6n«+27n+29  29
86 6(n+l)(n+2)(n+3)' ^"~36*
13. fir^=^. ^00=2. 14. i(»+l)(n+2)(4n+8)J.
15. jln(n+l)(n+2)(8w»+lln+l).
16. na*+n(nl)a6+g(»l)n(2nl)6».
17. wa8+n(nl)a26 + (nl)n(2nl)a6»+ln«(nl)26».
18. 5to(4»«1). 10. l»(16n«12nl).
9 9
28. gn(»+l)(»+2). 84. n(n + l)(4nl).
85. naft5n(nl)(a+6)+gn(nl)(2nl).
2»+i 1
87. (i) s1. (li) 1; rrj^«
^' n+2 ^ ' (n+l)2*
/• x o 2 /2V ... 5 * 5 /5\»
("^) ^i+iU) • <^") i 2(nH)(nh2) V7) *
( \ ^ 8 /3\» . .. 6 /6\»
^^^ 2"'(n+l)(n+2)VV ' ^^^ (n + 1) (n+2) V?/ *
J
ANSWERS TO THE EXAMPLES. 619
EXAMPLES XXXIV.
4.7.10...(3n+4) 2.5. 8...(8n+2)
• ^^ 2.6.8...(3n + 2) " ^ ^ 4.7.10...(3n + l)"' *
5.7...(2n + 3) . r 13.15...(2n+ll))
^""^^ 8.10...(2»+6)' ^ ^'^^ ^^r 14.16...(2n+12)f' ^^•
2. (i) 2 + 3(nl)(n2);>+n(nl)(n2).
(ii) 7n(nl)(n2); ^n(n+l)in(nl) (n2).
(iii) 2»+i»2; 2*^^ll(» + 2)(n + 3).
(iv) 2*+in(n + l)n; 2»+34in (n+1) (ti + 2)in(n + l).
(v) ^ti(n+l)(n+2)(n+3); j^n(n+l)(n+2) (n+B) (n+^).
(vi) (n2)(nl)»(n + l) + (»l)titi+2;
i(n2)(»l)n{n+l)(n+2)+(».l)»(n+l)in(n + l)+2n.
^ „. 24aj ,.,. l2a; ,„.. l6af
l.4a.+a;a* ^"' l5a:+4xa* ^"'^ l12a; + 32x>*
/• \ 15 + 0? 19g' , . l + x
1614x86a;«42«»' ^ ' (1aj)**
4. (i) 2«+i2; 2*«2n4.
(ii) l{8+ll(.4)l};^ + ^g(4)^
(iii) l{3»(l)*}; g {3»+i  8} when n is even, and i {3*+i  1}
when n is odd.
•• ^{(1+>/5)*+(1n/5)"}. • a=l, 6=4, c=l, d=0.
23x«« ^ 1
(1  «)« (1  2a;) ' (la;)a(l2a;)*
620 ANSWERS TO THE EXAMPLES.
24. l(^ljlog(la?). 25. ^T_^^4 ' »0 *<!•
nvergei
Convergent if 7 > a + /3, Divergent if 7 < a + /S.
82. (i) Divergent, (ii) Divergent, (iii) Convergent, (iv) Convergent.
(v)Co ■
EXAMPLES XX^VL
n+1
lO &i»iP'h  (Mh + ft Vi»l) PVi  <»nfti» 1 (ftn&nl + ^n) A3
EXAMPLES XXXVn.
I. (i) 4+i (ii) 11+i 7 T SH
^^ 8+ ^ ' 1 + 4+1 + 22+
<^) ^+i+i+I+^+
.1111111111^
(IV) t>+i^i_^3^i^g^j;^3_j.l^j_j.l2+
•• « s/v (") f(^+>/37). (in) ^(28V30).
16. 5(n«+8n). 26. «"J.
EXAMPLES XXXVm.
6. 733. 10. 3, 7, 9, 11, 13, 19. 18. 504f»6.
EXAMPLES XL.
1. (i) aj=2, y=:3. (u) «=1, y=10; x=U, y=2.
(iii) aj=4, y=8; a;=13, y=l.
(iv) 696, 3; 626, 18; 664, 33; ; 67, 138.
ANSWERS TO THE BXAUPLES.
9. (i) X!=i + lSm, y=l + 'Jm. (iij i = 9 + llin, ]/=7 + ftn.
(iii) «!=16m7.y = 17m10. (iv) a: = 64 + 69n., y=M + 49»».
4. (i) S, 1, 2; 5, 3, I; 3, 4, I.
(ii) 1, 31, 1; 5, 14, I; 9, 7, 1; S, 13, 2; 7, 6, 3; 6,6,8; 8,4,4;
1,13,3; 1,3,5.
(Ui) 2,8,3. H 8,38,60; 19,44,36; 3O,S0,30; 41,66,6.
(i) 1826, 2; 441. 8; 101, 8; 77, 10; 83, 21; 36, 27; 6, 113; 1, 3B3.
(U) S,S. (iii) 8,6. (ii) 6,1; 18,14.
195, 121; 63, 264.
8. ». 20. la 8. It. £3. 14<. 6d., £4. 6i. Gd.
11, 12, 16, 34, 36. 1«. 15, 55; 26, 66; 3
EXAMPLES XLL
86 31
8. 8n* + 6n+>p«noe. ■•
EXAMPLES XLIU.
I. ii)q. (ii)373. (iii) ap. (i»)8«. {v)2p'. (vi) Sj.
(Tii) 2p>. (Viii) 3i^. (ix) ^. (I) piq. (ri) p/29.
(xii) p'/(145>+p'.
B. (i) 3p"16pj + 64r. (ii) (g"
t. (i) 2S, 34. (ii) 44, 168.
B. {i)Pi'2p>, (ii) 3p,ft
H (8J'.P,J>i'8P.)(P..*
622 ANSWERS TO THE EXAMPLES.
6. «»10x8+31aj31=0. 7. 6. 8. (i) a^qa:^+prxr^=0,
(ii) as» + 2pxs+(pa + g)yr+pg=0.
(iii) a;»(rpg)+a^»(3r~2pg+j)«)+x(3ri)g)+r=0.
(iv) afi2qx^+{q^{'pr)x+i^pqr=:0.
2r
(v) Eliminate x between given equation and y={p+x^i — .
X
(yi) y*{Sqp*)y^+{Sq*'qp^)y+rp^q^^O.
•. (i) Substitute  {y +p) for x,
(ii) Substitute  « (y +p) for x,
o
(iii) Substitute 5 (l>*  2g  y) for oc".
EXAMPLES XLIV.
3. , ±V2±V6. 4 ±V2±>/^, (1±n/7).
5. a;*16a:2+4=0. 6. aJ*+2aj"+25=0.
8. 1±V^, 1^2*7^. O. 2g»9pgr + 271^=0.
10. j)«  4i)g + 8r = and {p^ + 4g) (36g  llp^)  1600s. = 0.
11. 6, 1, 3. la. 4, 1, 2, 5.
13. (i) +4, 4. (ii) 3. (iii) . 14. 8, ±^.
10. ±,^2, 2±^7. 16. r»pgr+p'8=0.
4
17. 3, 7, 10. 18. 3,9, g.
20. (i) r2)»«=0. (ii) p««+2»=4g«.
EXAMPLES XLV.
I. (i)»^.4 (ii)3,3, ±2V^. (iii) , , J, ?.
(iv) 4, 4, 4,  i . 2. (6c  ad)a= 4 (J^  ac) (c*  bd).
9, (i) y»6gy«+4(4i)r«)y8(2p2,«3g«+2r»).
(ii) (y6g)» + 6g(y6g)«+4(^«)(y6g)+8(2p«»3gir+2r«)=0.
10. (y  16p* + 24g)8  24g (y  16p2 + 24g)«
+ 64 (4r  «) (y  16^^^ + 24g)  512 (2p3«  8^ + 2t3) =0,
ANSWERS TO THE EXAMPLES. 623
EXAMPLES XLVI.
1. (i) ~6, ~w4w', w^— 4w, i.e. 6, ^ =t ^ ^^  3.
(ii) _ 4, _ 0,  3a^, _ 0,2 _ 3w. (iii) 10, 2w + 8«3, 2u^ + 8«.
(iv) 8, w + 7w^ w2^.7t^
(v) 2094...,  1703.. .« 391.. .w2, l703...«2391...«.
(vi) 30913, 2'1699w+9214««, 21699w2+9214w.
a. (i) 1, 3, l±2i. (ii) 1±V2, 1=4=^^5.
(iii) 3±V^, 3±V^. (iv) 2, , i(l±y::i5).
8. (i) One real root between  3 and  4.
(ii) One between 7 and 6, one between 1 and 2, and one
between 5 and 6.
(iii) One between and 1, and one between 1 and 2.
(iv) One between 2 and 3, one between and  1, and one between
 4 and  5.
(y) One between 2 and 3, and one between  3 and  4.
8. (i) 13670, 16139. (ii) 41891. (iii) 4679, 16527, 38793.
(iv) 22318. (V) 21624, 24142. (vi) 11487.
9. (i) 3, 3, 4, 4. (ii) 3, 3, 3^^^.
11. 1,3, s, ±V^^.
(SavAxSasst: ■
t C. J, OLAT, M.A. A aOHB,